JEE Main 2024 — Vectors & 3D Geometry

A line passes through $A(4,-6,-2)$ and $B(16,-2,4)$. The point $P(a,b,c)$ where $a,b,c$ are non-negative integers, on the line $AB$ lies at a distance of $21$ units, from the point $A$. The distance between the points $P(a,b,c)$ and $Q(4,-12,3)$ is equal to ______.

A line with direction ratio $2,1,2$ meets the lines $x=y+2=z$ and $x+2=2y=2z$ respectively at the point $P$ and $Q$. if the length of the perpendicular from the point $(1,2,12)$ to the line $\mathrm{PQ}$ is $l$, then ${l}^{2}$ is

Between the following two statements: Statement I : Let $\vec{a}=\hat{i}+2 \hat{j}-3 \hat{k}$ and $\vec{b}=2 \hat{i}+\hat{j}-\hat{k}$. Then the vector $\vec{r}$ satisfying $\vec{a} \times \vec{r}=\vec{a} \times \vec{b}$ and $\vec{a} \cdot \vec{r}=0$ is of magnitude $\sqrt{10}$. Statement II : In a triangle $A B C, \cos 2 A+\cos 2 B+\cos 2 C \geq-\frac{3}{2}$.

Consider a line $\mathrm{L}$ passing through the points $\mathrm{P}(1,2,1)$ and $\mathrm{Q}(2,1,-1)$. If the mirror image of the point $\mathrm{A}(2,2,2)$ in the line $\mathrm{L}$ is $(\alpha, \beta, \gamma)$, then $\alpha+\beta+6 \gamma$ is equal to _______

Consider a $\Delta ABC$ where $A(1,3,2),B(-2,8,0)$ and $C(3,6,7)$. If the angle bisector of $\angle BAC$ meets the line $BC$ at $D$, then the length of the projection of the vector $\vec{AD}$ on the vector $\vec{AC}$ is:

Consider the line $L$ passing through the points $(1,2,3)$ and $(2,3,5)$. The distance of the point $\left(\frac{11}{3}, \frac{11}{3}, \frac{19}{3}\right)$ from the line $\mathrm{L}$ along the line $\frac{3 x-11}{2}=\frac{3 y-11}{1}=\frac{3 z-19}{2}$ is equal to

Consider three vectors $\vec{a}, \vec{b}, \vec{c}$. Let $|\vec{a}|=2,|\vec{b}|=3$ and $\vec{a}=\vec{b} \times \vec{c}$. If $\alpha \in\left[0, \frac{\pi}{3}\right]$ is the angle between the vectors $\vec{b}$ and $\vec{c}$, then the minimum value of $27|\vec{c}-\vec{a}|^2$ is equal to:

For $\lambda>0$, let $\theta$ be the angle between the vectors $\vec{a}=\hat{i}+\lambda \hat{j}-3 \hat{k}$ and $\vec{b}=3 \hat{i}-\hat{j}+2 \hat{k}$. If the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are mutually perpendicular, then the value of (14 cos $\theta)^2$ is equal to

If $\vec{a}=\hat{i}+2\hat{j}+\hat{k},\vec{b}=3(\hat{i}-\hat{j}+\hat{k})$ and $\vec{c}$ be the vector such that $\vec{a}\times \vec{c}=\vec{b}$ and $\vec{a}\cdot \vec{c}=3$, then $\vec{a}\cdot ((\vec{c}\times \vec{b})-\vec{b}-\vec{c})$ is equal to

If ${d}_{1}$ is the shortest distance between the lines $x+1=2y=-12z,x=y+2=6z-6$ and ${d}_{2}$ is the shortest distance between the lines $\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5},\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}$, then the value of $\frac{32\sqrt{3}{d}_{1}}{{d}_{2}}$ is :

If the line $\frac{2-x}{3}=\frac{3 y-2}{4 \lambda+1}=4-z$ makes a right angle with the line $\frac{x+3}{3 \mu}=\frac{1-2 y}{6}=\frac{5-z}{7}$, then $4 \lambda+9 \mu$ is equal to :

If the mirror image of the point $P(3,4,9)$ in the line $\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-2}{1}$ is $(\alpha ,\beta ,\gamma ),$ then $14(\alpha +\beta +\gamma )$ is:

If the shortest distance between the lines $\frac{x+2}{2}=\frac{y+3}{3}=\frac{z-5}{4}$ and $\frac{x-3}{1}=\frac{y-2}{-3}=\frac{z+4}{2}$ is $\frac{38}{3 \sqrt{5}} \mathrm{k}$, and $\int_0^{\mathrm{k}}\left[x^2\right] \mathrm{d} x=\alpha-\sqrt{\alpha}$, where $[x]$ denotes the greatest integer function, then $6 \alpha^3$ is equal to________

If the shortest distance between the lines $\frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3}$ and $\frac{x-\lambda }{2}=\frac{y+1}{4}=\frac{z-2}{-5}$ is $\frac{6}{\sqrt{5}}$, then the sum of all possible values of $\lambda$ is :

If the shortest distance between the lines $\frac{x-\lambda}{2}=\frac{y-4}{3}=\frac{z-3}{4}$ and $\frac{x-2}{4}=\frac{y-4}{6}=\frac{z-7}{8}$ is $\frac{13}{\sqrt{29}}$, then a value of $\lambda$ is :

If the shortest distance between the lines $\frac{x-\lambda}{3}=\frac{y-2}{-1}=\frac{z-1}{1}$ and $\frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4}$ is $\frac{44}{\sqrt{30}}$, then the largest possible value of $|\lambda|$ is equal to _________

If the shortest distance between the lines $\frac{x-\lambda }{-2}=\frac{y-2}{1}=\frac{z-1}{1}$ and $\frac{x-\sqrt{3}}{1}=\frac{y-1}{-2}=\frac{z-2}{1}$ is $1$, then the sum of all possible values of $\lambda$ is

If the shortest distance between the lines $\begin{array}{ll} L_1: \vec{r}=(2+\lambda) \hat{i}+(1-3 \lambda) \hat{j}+(3+4 \lambda) \hat{k}, & \lambda \in \mathbb{R} \\ L_2: \vec{r}=2(1+\mu) \hat{i}+3(1+\mu) \hat{j}+(5+\mu) \hat{k}, \quad \mu \in \mathbb{R} \end{array}$ is $\frac{m}{\sqrt{n}}$, where $\operatorname{gcd}(m, n)=1$, then the value of $m+n$ equals

The image of the point (1,2,3) in the plane x+y+z=9 is:

The projection of vector a=2i-3j+6k on vector b=i+2j+2k is:

Let a line passing through the point $(-1,2,3)$ intersect the lines ${L}_{1}:\frac{x-1}{3}=\frac{y-2}{2}=\frac{z+1}{-2}$ at $M(\alpha ,\beta ,\gamma )$ and ${L}_{2}:\frac{x+2}{-3}=\frac{y-2}{-2}=\frac{z-1}{4}$ at $N(a,b,c)$. Then the value of $\frac{(\alpha +\beta +\gamma {)}^{2}}{(a+b+c{)}^{2}}$ equals ________________.

Let a unit vector $\hat{u}=x\hat{i}+y\hat{j}+z\hat{k}$ make angles $\frac{\pi }{2},\frac{\pi }{3}$ and $\frac{2\pi }{3}$ with the vectors $\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{k},\frac{1}{\sqrt{2}}\hat{j}+\frac{1}{\sqrt{2}}\hat{k}$ and $\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}$ respectively. If $\vec{v}=\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}+\frac{1}{\sqrt{2}}\hat{k}$, then \(|\hat{u}-\vec{v}|^2\) is equal to

Let a unit vector which makes an angle of $60^{\circ}$ with $2 \hat{i}+2 \hat{j}-\hat{k}$ and angle $45^{\circ}$ with $\hat{i}-\hat{k}$ be $\overrightarrow{\mathrm{C}}$. Then $\overrightarrow{\mathrm{C}}+\left(-\frac{1}{2} \hat{i}+\frac{1}{3 \sqrt{2}} \hat{j}-\frac{\sqrt{2}}{3} \hat{k}\right)$ is :

Let $\vec{a}=2 \hat{i}-3 \hat{j}+4 \hat{k}, \vec{b}=3 \hat{i}+4 \hat{j}-5 \hat{k}$ and a vector $\vec{c}$ be such that $\vec{a} \times(\vec{b}+\vec{c})+\vec{b} \times \vec{c}=\hat{i}+8 \hat{j}+13 \hat{k}$. If $\vec{a} \cdot \vec{c}=13$, then $(24-\vec{b} \cdot \vec{c})$ is equal to_______

Let $\vec{a}=3\hat{i}+2\hat{j}+\hat{k},\vec{b}=2\hat{i}-\hat{j}+3\hat{k}$ and $\vec{c}$ be a vector such that $(\vec{a}+\vec{b})\times \vec{c}=2(\vec{a}\times \vec{b})+24\hat{j}-6\hat{k}$ and $(\vec{a}-\vec{b}+\hat{i}).\vec{c}=-3$. Then ${|\vec{c}|}^{2}$ is equal to _______.

Let $\vec{a}=4 \hat{i}-\hat{j}+\hat{k}, \vec{b}=11 \hat{i}-\hat{j}+\hat{k}$ and $\vec{c}$ be a vector such that $(\vec{a}+\vec{b}) \times \vec{c}=\vec{c} \times(-2 \vec{a}+3 \vec{b})$. If $(2 \vec{a}+3 \vec{b}) \cdot \vec{c}=1670$, then $|\vec{c}|^2$ is equal to :

Let $\overrightarrow{\mathrm{a}}=\hat{i}-3 \hat{j}+7 \hat{k}, \overrightarrow{\mathrm{b}}=2 \hat{i}-\hat{j}+\hat{k}$ and $\overrightarrow{\mathrm{c}}$ be a vector such that $(\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=3(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}})$. If $\vec{a} \cdot \vec{c}=130$, then $\vec{b} \cdot \vec{c}$ is equal to _______

Let $A(2,3,5)$ and $C(-3,4,-2)$ be opposite vertices of a parallelogram $ABCD$ if the diagonal $\vec{BD}=\hat{i}+2\hat{j}+3\hat{k}$ then the area of the parallelogram is equal to

Let $Q$ and $R$ be the feet of perpendiculars from the point $P(a,a,a)$ on the lines $x=y,z=1$ and $x=-y,z=-1$ respectively. If $\angle QPR$ is a right angle, then $12{a}^{2}$ is equal to ________

Let $P(3,2,3),Q(4,6,2)$ and $R(7,3,2)$ be the vertices of $\Delta \mathrm{PQR}$. Then, the angle $\angle \mathrm{QPR}$ is

Let $\vec{a}=9 \hat{i}-13 \hat{j}+25 \hat{k}, \vec{b}=3 \hat{i}+7 \hat{j}-13 \hat{k}$ and $\vec{c}=17 \hat{i}-2 \hat{j}+\hat{k}$ be three given vectors. If $\vec{r}$ is a vector such that $\vec{r} \times \vec{a}=(\vec{b}+\vec{c}) \times \vec{a}$ and $\vec{r} \cdot(\vec{b}-\vec{c})=0$, then $\frac{|593 \vec{r}+67 \vec{a}|^2}{(593)^2}$ is equal to___________

Let ${L}_{1}:\vec{r}=(\hat{i}-\hat{j}+2\hat{k})+\lambda (\hat{i}-\hat{j}+2\hat{k}),\lambda \in R$, ${L}_{2}:\vec{r}=(\hat{j}-\hat{k})+\mu (3\hat{i}+\hat{j}+p\hat{k}),\mu \in R$ and ${L}_{3}:\vec{r}=\delta (l\hat{i}+m\hat{j}+n\hat{k}),\delta \in R$ be three lines such that ${L}_{1}$ is perpendicular to ${L}_{2}$ and ${L}_{3}$ is perpendicular to both ${L}_{1}$ and ${L}_{2}$. Then the point which lies on ${L}_{3}$ is

Let $\vec{a},\vec{b}$ and $\vec{c}$ be three non-zero vectors such that $\vec{b}$ and $\vec{c}$ are non-collinear if $\vec{a}+5\vec{b}$ is collinear with $\vec{c},\vec{b}+6\vec{c}$ is collinear with $\vec{a}$ and $\vec{a}+\alpha \vec{b}+\beta \vec{c}=\vec{0}$, then $\alpha +\beta$ is equal to

Let $\vec{a}=3\hat{i}+\hat{j}-2\hat{k},\vec{b}=4\hat{i}+\hat{j}+7\hat{k}$ and $\vec{c}=\hat{i}-3\hat{j}+4\hat{k}$ be three vectors. If a vectors $\vec{p}$ satisfies $\vec{p}\times \vec{b}=\vec{c}\times \vec{b}$ and $\vec{p}\cdot \vec{a}=0$, then $\vec{p}\cdot (\hat{i}-\hat{j}-\hat{k})$ is equal to

Let $\overrightarrow{\mathrm{a}}=\hat{i}+2 \hat{j}+3 \hat{k}, \overrightarrow{\mathrm{b}}=2 \hat{i}+3 \hat{j}-5 \hat{k}$ and $\overrightarrow{\mathrm{c}}=3 \hat{i}-\hat{j}+\lambda \hat{k}$ be three vectors. Let $\overrightarrow{\mathrm{r}}$ be anit vector along $\vec{b}+\vec{c}$. If $\vec{r} \cdot \vec{a}=3$, then $3 \lambda$ is equal to:

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k},\vec{b}=-\hat{i}-8\hat{j}+2\hat{k}$ and $\vec{c}=4\hat{i}+{c}_{2}\hat{j}+{c}_{3}\hat{k}$ be three vectors such that $\vec{b}\times \vec{a}=\vec{c}\times \vec{a}$. If the angle between the vector $\vec{c}$ and the vector $3\hat{i}+4\hat{j}+\hat{k}$ is $\theta$, then the greatest integer less than or equal to ${\mathrm{tan}}^{2}\theta$ is:

Let $\overrightarrow{\mathrm{a}}=2 \hat{i}+5 \hat{j}-\hat{k}, \overrightarrow{\mathrm{b}}=2 \hat{i}-2 \hat{j}+2 \hat{k}$ and $\overrightarrow{\mathrm{c}}$ be three vectors such that $(\vec{c}+\hat{i}) \times(\vec{a}+\vec{b}+\hat{i})=\vec{a} \times(\vec{c}+\hat{i})$. If $\vec{a} \cdot \vec{c}=-29$, then $\vec{c} \cdot(-2 \hat{i}+\hat{j}+\hat{k})$ is equal to:

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{a}|=1,|\vec{b}|=4$ and $\vec{a}\cdot \vec{b}=2$. If $\vec{c}=(2\vec{a}\times \vec{b})-3\vec{b}$ and the angle between $\vec{b}$ and$\vec{c}$ is $\alpha$, then $192{\mathrm{sin}}^{2}\alpha$ is equal to _________

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{b}|=1$ and $|\vec{b}\times \vec{a}|=2$ Then $|(\vec{b}\times \vec{a})-\vec{b}{|}^{2}$ is equal to

Let $\vec{a}={a}_{i}\hat{i}+{a}_{2}\hat{j}+{a}_{3}\hat{k}$ and $\vec{b}={b}_{1}\hat{i}+{b}_{2}\hat{j}+{b}_{3}\hat{k}$ be two vectors such that $|\vec{a}|=1;\vec{a}\cdot \vec{b}=2$ and $|\vec{b}|=4$. If $\vec{c}=2(\vec{a}\times \vec{b})-3\vec{b}$, then the angle between $\vec{b}$ and $\vec{c}$ is equal to :

Let $\overrightarrow{\mathrm{a}}=6 \hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{\mathrm{b}}=\hat{i}+\hat{j}$. If $\overrightarrow{\mathrm{c}}$ is a is vector such that $|\overrightarrow{\mathrm{c}}| \geq 6, \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=6|\overrightarrow{\mathrm{c}}|,|\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}|=2 \sqrt{2}$ and the angle between $\vec{a} \times \vec{b}$ and $\vec{c}$ is $60^{\circ}$, then $|(\vec{a} \times \vec{b}) \times \vec{c}|$ is equal to:

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}+4 \hat{j}-5 \hat{k}$ and $\vec{c}=x \hat{i}+2 \hat{j}+3 \hat{k}, x \in \mathbb{R}$. If $\vec{d}$ is the unit vector in the direction of $\vec{b}+\vec{c}$ such that $\vec{a} \cdot \vec{d}=1$, then $(\vec{a} \times \vec{b}) \cdot \vec{c}$ is equal to

Let $\vec{a}=-5\hat{i}+\hat{j}-3\hat{k},\vec{b}=\hat{i}+2\hat{j}-4\hat{k}$ and $\vec{c}=(((\vec{a}\times \vec{b})\times \hat{i})\times \hat{i})\times \hat{i}$. Then $\vec{c}\cdot (-\hat{i}+\hat{j}+\hat{k})$ is equal to

Let $\vec{\mathrm{OA}}=\vec{a},\vec{\mathrm{OB}}=12\vec{a}+4\vec{b}$ and $\vec{\mathrm{OC}}=\vec{b}$, where $O$ is the origin. If $S$ is the parallelogram with adjacent sides $\mathrm{OA}$ and $\mathrm{OC}$, then $\frac{\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{quadrilateral}\mathrm{OABC}}{\mathrm{area}\mathrm{of}S}$ is equal to _____

Let $\overrightarrow{O A}=2 \vec{a}, \overrightarrow{O B}=6 \vec{a}+5 \vec{b}$ and $\overrightarrow{O C}=3 \vec{b}$, where $O$ is the origin. If the area of the parallelogram with adjacent sides $\overrightarrow{\mathrm{OA}}$ and $\overrightarrow{\mathrm{OC}}$ is 15 sq. units, then the area (in sq. units) of the quadrilateral $\mathrm{OABC}$ is equal to :

Let $P(x, y, z)$ be a point in the first octant, whose projection in the $x y$-plane is the point $Q$. Let $O P=\gamma$; the angle between $O Q$ and the positive $x$-axis be $\theta$; and the angle between $O P$ and the positive $z$-axis be $\phi$, where $O$ is the origin. Then the distance of $P$ from the $x$-axis is

Let $\mathrm{ABC}$ be a triangle of area $15 \sqrt{2}$ and the vectors $\overrightarrow{\mathrm{AB}}=\hat{i}+2 \hat{j}-7 \hat{k}, \overrightarrow{\mathrm{BC}}=\mathrm{a} \hat{i}+\mathrm{b} \hat{j}+\mathrm{ck}$ and $\overrightarrow{\mathrm{AC}}=6 \hat{i}+\mathrm{d} \hat{j}-2 \hat{k}, \mathrm{~d}>0$. Then the square of the length of the largest side of the triangle $\mathrm{ABC}$ is _______

Let $PQR$ be a triangle with $R(-1,4,2)$. Suppose $M(2,1,2)$ is the mid point of $PQ$. The distance of the centroid of $\Delta PQR$ from the point of intersection of the line $\frac{x-2}{0}=\frac{y}{2}=\frac{z+3}{-1}$ and $\frac{x-1}{1}=\frac{y+3}{-3}=\frac{z+1}{1}$ is

Let $(\alpha ,\beta ,\gamma )$ be mirror image of the point $(2,3,5)$ in the line $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$. Then $2\alpha +3\beta +4\gamma$ is equal to

Let $\mathrm{d}$ be the distance of the point of intersection of the lines $\frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}$ and $\frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}$ from the point $(7,8,9)$. Then $\mathrm{d}^2+6$ is equal to :

Let $(\alpha ,\beta ,\gamma )$ be the foot of perpendicular from the point $(1,2,3)$ on the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$. then $19(\alpha +\beta +\gamma )$ is equal to :

Let $\mathrm{P}(\alpha, \beta, \gamma)$ be the image of the point $\mathrm{Q}(3,-3,1)$ in the line $\frac{x-0}{1}=\frac{y-3}{1}=\frac{z-1}{-1}$ and $\mathrm{R}$ be the point $(2,5,-1)$. If the area of the triangle $P Q R$ is $\lambda$ and $\lambda^2=14 K$, then $K$ is equal to :

Let $(\alpha, \beta, \gamma)$ be the image of the point $(8,5,7)$ in the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{5}$. Then $\alpha+\beta+\gamma$ is equal to :

Let $\mathrm{P}(\alpha, \beta, \gamma)$ be the image of the point $\mathrm{Q}(1,6,4)$ in the line $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$. Then $2 \alpha+\beta+\gamma$ is equal to_______

Let $O$ be the origin, and $M$ and $N$ be the points on the lines $\frac{x-5}{4}=\frac{y-4}{1}=\frac{z-5}{3}$ and $\frac{x+8}{12}=\frac{y+2}{5}=\frac{z+11}{9}$ respectively such that $MN$ is the shortest distance between the given lines. Then $\vec{OM}\cdot \vec{ON}$ is equal to _________.

Let $O$ be the origin and the position vector of $A$ and $B$ be $2\hat{i}+2\hat{j}+\hat{k}$ and $2\hat{i}+4\hat{j}+4\hat{k}$ respectively. If the internal bisector of $\angle AOB$ meets the line $AB$ at $C$, then the length of $OC$ is

Let $P$ be the point $(10,-2,-1)$ and $Q$ be the foot of the perpendicular drawn from the point $R(1,7,6)$ on the line passing through the points $(2,-5,11)$ and $(-6,7,-5)$. Then the length of the line segment $P Q$ is equal to ________

Let $\mathrm{P}$ be the point of intersection of the lines $\frac{x-2}{1}=\frac{y-4}{5}=\frac{z-2}{1}$ and $\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-3}{2}$. Then, the shortest distance of P from the line $4 x=2 y=z$ is

Let $P\text{and}Q$ be the points on the line $\frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2}$ which are at a distance of $6$ units from the point $R(1,2,3)$. If the centroid of the triangle $PQR$ is $(\alpha ,\beta ,\gamma ),$ then ${\alpha }^{2}+{\beta }^{2}+{\gamma }^{2}$ is:

Let $\vec{a}=\hat{i}+\alpha \hat{j}+\beta \hat{k},\alpha ,\beta \in R$. Let a vector $\vec{b}$ be such that the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi }{4}$ and ${|\vec{b}|}^{2}=6$, If $\vec{a}\cdot \vec{b}=3\sqrt{2}$, then the value of $({\alpha }^{2}+{\beta }^{2})|\vec{a}\times \vec{b}{|}^{2}$ is equal to

Let the image of the point $(1,0,7)$ in the line $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$ be the point $(\alpha ,\beta ,\gamma )$. Then which one of the following points lies on the line passing through $(\alpha ,\beta ,\gamma )$ and making angles $\frac{2\pi }{3}$ and $\frac{3\pi }{4}$ with $y-$axis and $z-$axis respectively and an acute angle with $x-$axis?

Let the line $\mathrm{L}$ intersect the lines $x-2=-y=z-1,2(x+1)=2(y-1)=z+1$ and be parallel to the line $\frac{x-2}{3}=\frac{y-1}{1}=\frac{z-2}{2}$. Then which of the following points lies on L?

Let the line of the shortest distance between the lines ${L}_{1}:\vec{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda (\hat{i}-\hat{j}+\hat{k})$ and ${L}_{2}:\vec{r}=(4\hat{i}+5\hat{j}+6\hat{k})+\mu (\hat{i}+\hat{j}-\hat{k})$ intersect ${L}_{1}$ and ${L}_{2}$ at $P$ and $Q$ respectively. If $(\alpha ,\beta ,\gamma )$ is the midpoint of the line segment $PQ$, then $2(\alpha +\beta +\gamma )$ is equal to___________

Let the point $(-1, \alpha, \beta)$ lie on the line of the shortest distance between the lines $\frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2}$ and $\frac{x+2}{-1}=\frac{y+6}{2}=\frac{z-1}{0}$. Then $(\alpha-\beta)^2$ is equal to___________

Let the point, on the line passing through the points $P(1,-2,3)$ and $Q(5,-4,7)$, farther from the origin and at distance of 9 units from the point $\mathrm{P}$, be $(\alpha, \beta, \gamma)$. Then $\alpha^2+\beta^2+\gamma^2$ is equal to :

Let the position vectors of the vertices $A,B$ and $C$ of a triangle be $2\hat{i}+2\hat{j}+\hat{k},\hat{i}+2\hat{j}+2\hat{k}$ and $2\hat{i}+\hat{j}+2\hat{k}$ respectively. Let ${l}_{1},{l}_{2}$ and ${l}_{3}$ be the lengths of perpendiculars drawn from the ortho centre of the triangle on the sides $\mathrm{AB},\mathrm{BC}$ and $\mathrm{CA}$ respectively, then ${l}_{1}^{2}+{l}_{2}^{2}+{l}_{3}^{2}$ equals :

Let $\overrightarrow{\mathrm{a}}=2 \hat{i}+\hat{j}-\hat{k}, \overrightarrow{\mathrm{b}}=((\overrightarrow{\mathrm{a}} \times(\hat{i}+\hat{j})) \times \hat{i}) \times \hat{i}$. Then the square of the projection of $\overrightarrow{\mathrm{a}}$ on $\overrightarrow{\mathrm{b}}$ is :

Let three vectors $\overrightarrow{\mathrm{a}}=\alpha \hat{i}+4 \hat{j}+2 \hat{k}, \overrightarrow{\mathrm{b}}=5 \hat{i}+3 \hat{j}+4 \hat{k}, \overrightarrow{\mathrm{c}}=x \hat{i}+y \hat{j}+z \hat{k}$ form a triangle such that $\vec{c}=\vec{a}-\vec{b}$ and the area of the triangle is $5 \sqrt{6}$. If $\alpha$ is a positive real number, then $|\vec{c}|^2$ is equal to:

Let $\vec{a}=2 \hat{i}+\alpha \hat{j}+\hat{k}, \vec{b}=-\hat{i}+\hat{k}, \vec{c}=\beta \hat{j}-\hat{k}$, where $\alpha$ and $\beta$ are integers and $\alpha \beta=-6$. Let the values of the ordered pair $(\alpha, \beta)$, for which the area of the parallelogram of diagonals $\vec{a}+\vec{b}$ and $\vec{b}+\vec{c}$ is $\frac{\sqrt{21}}{2}$, be $\left(\alpha_1, \beta_1\right)$ and $\left(\alpha_2, \beta_2\right)$. Then $\alpha_1^2+\beta_1^2-\alpha_2 \beta_2$ is equal to

The distance of the point $Q(0,2,–2)$ form the line passing through the point $P(5,–4,3)$ and perpendicular to the lines $\vec{r}=(-3\hat{i}+2\hat{k})+\lambda (2\hat{i}+3\hat{j}+5\hat{k}),\lambda \in \mathbb{R}$ and $\vec{r}=(\hat{i}-2\hat{j}+\hat{k})+\mu (-\hat{i}+3\hat{j}+2\hat{k}),\mu \in \mathbb{R}$ is

The distance, of the point $(7,-2,11)$ from the line $\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}$ along the line $\frac{x-5}{2}=\frac{y-1}{-3}=\frac{z-5}{6}$, is :

The least positive integral value of $\alpha$, for which the angle between the vectors $\alpha \hat{i}-2\hat{j}+2\hat{k}$ and $\alpha \hat{i}+2\alpha \hat{j}-2\hat{k}$ is acute, is _____.

The lines $\frac{x-2}{2}=\frac{y}{-2}=\frac{z-7}{16}$ and $\frac{x+3}{4}=\frac{y+2}{3}=\frac{z+2}{1}$ intersect at the point $P$. If the distance of $P$ from the line $\frac{x+1}{2}=\frac{y-1}{3}=\frac{z-1}{1}$ is $l$, then $14{l}^{2}$ is equal to _____.

The position vectors of the vertices $A,B$ and $C$ of a triangle are $2\hat{i}-3\hat{j}+3\hat{k},2\hat{i}+2\hat{j}+3\hat{k}$ and $-\hat{i}+\hat{j}+3\hat{k}$ respectively. Let $l$ denotes the length of the angle bisector $\mathrm{AD}$ of $\angle \mathrm{BAC}$ where $D$ is on the line segment $\mathrm{BC}$, then $2{l}^{2}$ equals :

The set of all $\alpha$, for which the vectors $\vec{a}=\alpha t \hat{i}+6 \hat{j}-3 \hat{k}$ and $\vec{b}=t \hat{i}-2 \hat{j}-2 \alpha t \hat{k}$ are inclined at an obtuse angle for all $t \in \mathbb{R}$, is

The shortest distance between lines ${L}_{1}$ and ${L}_{2}$, where ${L}_{1}:\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+4}{2}$ and ${L}_{2}$ is the line passing through the points $A(-4,4,3),B(-1,6,3)$ and perpendicular to the line $\frac{x-3}{-2}=\frac{y}{3}=\frac{z-1}{1}$, is

The shortest distance between the lines $\frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5}$ and $\frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3}$ is

The shortest distance between the lines $\frac{x-3}{4}=\frac{y+7}{-11}=\frac{z-1}{5}$ and $\frac{x-5}{3}=\frac{y-9}{-6}=\frac{z+2}{1}$ is:

The square of the distance of the image of the point $(6,1,5)$ in the line $\frac{x-1}{3}=\frac{y}{2}=\frac{z-2}{4}$, from the origin is _________