JEE Main 2026 — Vectors & 3D Geometry

If the distances of the point $(1,2, a)$ from the line $\frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}$ along the lines $\mathrm{L}_{1}: \frac{x-1}{3}=\frac{y-2}{4}=\frac{z-a}{b}$ and $\mathrm{L}_{2}: \frac{x-1}{1}=\frac{y-2}{4}=\frac{z-a}{c}$ are equal, then $a+b+c$ is equal to

If the point of intersection of the lines $\dfrac{x+1}{3} = \dfrac{y+a}{5} = \dfrac{z+b+1}{7}$ and $\dfrac{x-2}{1} = \dfrac{y-b}{4} = \dfrac{z-2a}{7}$ lies on $xy$-plane, then the value of $a + b$ is :

The volume of the parallelepiped formed by vectors a=i+2j-k, b=2i-j+3k, c=3i+j+2k is:

If the distance of the point $\mathrm{P}(43, \alpha, \beta), \beta<0$, from the line $\overrightarrow{\mathrm{r}}=4 \hat{i}-\hat{k}+\mu(2 \hat{i}+3 \hat{k}), \mu \in \mathbf{R}$ along a line with direction ratios $3,-1,0$ is $13 \sqrt{10}$, then $\alpha^{2}+\beta^{2}$ is equal to $\_\_\_\_$

Let $\vec{a}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}, \vec{b}=\hat{\mathrm{i}}+\hat{\mathrm{j}}$ and $\vec{c}=\vec{a} \times \vec{b}$. Let $\vec{d}$ be a vector such that $|\vec{d}-\vec{a}|=\sqrt{11},|\vec{c} \times \vec{d}|=3$ and the angle between $\vec{c}$ and $\vec{d}$ is $\frac{\pi}{4}$. Then $\vec{a} \cdot \vec{d}$ is equal to

Let the direction cosines of two lines satisfy the equations : $4 l+m-n=0$ and $2 m n+10 n l+3 l m=0$. Then the cosine of the acute angle between these lines is :

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

Let $\vec{a}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}, \vec{b}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}, \vec{c}=\lambda \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\vec{v}=\vec{a} \times \vec{b}$. If $\vec{v} \cdot \vec{c}=11$ and the length of the projection of $\vec{b}$ on $\vec{c}$ is $p$, then $9 p^{2}$ is equal to

Let the lines $\mathrm{L}_{1}: \vec{r}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}+\lambda(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}), \lambda \in \mathbb{R}$ and $\mathrm{L}_{2}: \vec{r}=(4 \hat{\mathrm{i}}+\hat{\mathrm{j}})+\mu(5 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}), \mu \in \mathbb{R}$, intersect at the point R. Let P and Q be the points lying on lines $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$, respectively, such that $|\overrightarrow{\mathrm{PR}}|=\sqrt{29}$ and $|\overrightarrow{\mathrm{PQ}}|=\sqrt{\frac{47}{3}}$. If the point P lies in the first octant, then $27(\mathrm{QR})^{2}$ is equal to

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

Let $\vec{a} = 2\hat{i} + 3\hat{j} + 3\hat{k}$ and $\vec{b} = 6\hat{i} + 3\hat{j} + 3\hat{k}$. Then the square of the area of the triangle with adjacent sides determined by the vectors $(2\vec{a} + 3\vec{b})$ and $(\vec{a} - \vec{b})$ is :

The shortest distance between the lines $\vec{r}=\left(\dfrac{1}{3}\hat{i}+2\hat{j}+\dfrac{8}{3}\hat{k}\right)+\lambda(2\hat{i}-5\hat{j}+6\hat{k})$ and $\vec{r}=\left(-\dfrac{2}{3}\hat{i}-\dfrac{1}{3}\hat{k}\right)+\mu(\hat{j}-\hat{k})$, $\lambda,\mu \in \mathbb{R}$, is:

For three unit vectors $\vec{a}, \vec{b}, \vec{c}$ satisfying $|\vec{a}-\vec{b}|^{2}+|\vec{b}-\vec{c}|^{2}+|\vec{c}-\vec{a}|^{2}=9$ and $|2 \vec{a}+k \vec{b}+k \vec{c}|=3$, the positive value of k is

Let $\vec{a} = 4\hat{i} - \hat{j} + 3\hat{k}$, $\vec{b} = 10\hat{i} + 2\hat{j} - \hat{k}$ and a vector $\vec{c}$ be such that $2(\vec{a}\times\vec{b}) + 3(\vec{b}\times\vec{c}) = \vec{0}$. If $\vec{a}\cdot\vec{c} = 15$, then $\vec{c}\cdot(\hat{i}+\hat{j}-3\hat{k})$ is equal to:

For a triangle ABC, let $\overrightarrow{\mathrm{p}}=\overrightarrow{\mathrm{BC}}, \overrightarrow{\mathrm{q}}=\overrightarrow{\mathrm{CA}}$ and $\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{BA}}$. If $|\overrightarrow{\mathrm{p}}|=2 \sqrt{3},|\overrightarrow{\mathrm{q}}|=2$ and $\cos \theta=\frac{1}{\sqrt{3}}$, where $\theta$ is the angle between $\vec{p}$ and $\vec{q}$, then $|\vec{p} \times(\vec{q}-3 \vec{r})|^{2}+3|\vec{r}|^{2}$ is equal to :

Let a line $L$ be perpendicular to both the lines $L_1: \dfrac{x+1}{3} = \dfrac{y+3}{5} = \dfrac{z+5}{7}$ and $L_2: \dfrac{x-2}{1} = \dfrac{y-4}{4} = \dfrac{z-6}{7}$. If $\theta$ is the acute angle between the lines $L$ and $L_3: \dfrac{x - \dfrac{8}{7}}{2} = \dfrac{y - \dfrac{4}{7}}{1} = \dfrac{z}{2}$, then $\tan\theta$ is equal to:

The sum of all values of $\alpha$, for which the shortest distance between the lines $\frac{x+1}{\alpha}=\frac{y-2}{-1}=\frac{z-4}{-\alpha}$ and $\frac{x}{\alpha}=\frac{y-1}{2}=\frac{z-1}{2 \alpha}$ is $\sqrt{2}$, is

If the image of the point $\mathrm{P}(1,2, a)$ in the line $\frac{x-6}{3}=\frac{y-7}{2}=\frac{7-\mathrm{z}}{2}$ is $\mathrm{Q}(5, b, \mathrm{c})$, then $a^{2}+b^{2}+c^{2}$ is equal to

Let $O$ be the origin, $\vec{OP} = \vec{a}$ and $\vec{OQ} = \vec{b}$. If $R$ is the point on $\vec{OP}$ such that $\vec{OP} = 5\vec{OR}$, and $M$ is the point such that $\vec{OQ} = 5\vec{RM}$, then $\vec{PM}$ is equal to :

Let $\vec{a} = \sqrt{7}\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{j} + 2\hat{k}$. If $\vec{r}$ is a vector such that $\vec{r} \times \vec{a} + \vec{a} \times \vec{b} = \vec{0}$ and $\vec{r} \cdot \vec{a} = 0$, then $|3\vec{r}|^2$ is equal to:

Let $P$ be a point in the plane of the vectors $\overrightarrow{A B}=3 \hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{A C}=\hat{i}-\hat{j}+3 \hat{k}$ such that $P$ is equidistant from the lines AB and AC. If $|\overrightarrow{\mathrm{AP}}|=\frac{\sqrt{5}}{2}$, then the area of the triangle ABP is:

Let $\vec{a}=2 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}$ and $\vec{b}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}$. If $\vec{c}$ is a vector such that $2(\vec{a} \times \vec{c})+3(\vec{b} \times \vec{c})=\overrightarrow{0}$ and $(\vec{a}-\vec{b}) \cdot \vec{c}=-97$, then $|\vec{c} \times \hat{\mathrm{k}}|^{2}$ is equal to

Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors such that $\vec{a} \times \vec{b}=2(\vec{a} \times \vec{c})$. If $|\vec{a}|=1,|\vec{b}|=4,|\vec{c}|=2$, and the angle between $\vec{b}$ and $\vec{c}$ is $60^{\circ}$, then $|\vec{a} \cdot \vec{c}|$ is equal to

Let $\vec{a}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=\lambda \hat{j}+2 \hat{k}, \lambda \in \boldsymbol{Z}$ be two vectors. Let $\vec{c}=\vec{a} \times \vec{b}$ and $\vec{d}$ be a vector of magnitude 2 in $y z$-plane. If $|\overrightarrow{\mathrm{c}}|=\sqrt{53}$, then the maximum possible value of $(\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{d}})^{2}$ is equal to :

Let $\overrightarrow{\mathrm{a}}=-\hat{i}+2 \hat{j}+2 \hat{k}, \overrightarrow{\mathrm{~b}}=8 \hat{i}+7 \hat{j}-3 \hat{k}$ and $\overrightarrow{\mathrm{c}}$ be vector such that $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}}$. If $\vec{c} \cdot(\hat{i}+\hat{j}+\hat{k})=4$, then $|\vec{a}+\vec{c}|^{2}$ is equal to :

Let a triangle $PQR$ be such that $P$ and $Q$ lie on the line $\dfrac{x+3}{8} = \dfrac{y-4}{2} = \dfrac{z+1}{2}$ and are at a distance of $6$ units from $R(1, 2, 3)$. If $(\alpha, \beta, \gamma)$ is the centroid of $\triangle PQR$, then $\alpha + \beta + \gamma$ is equal to :

Let $\mathrm{Q}(\mathrm{a}, \mathrm{b}, \mathrm{c})$ be the image of the point $\mathrm{P}(3,2,1)$ in the line $\frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}$. Then the distance of Q from the line $\frac{x-9}{3}=\frac{y-9}{2}=\frac{z-5}{-2}$ is

Let a line L passing through the point $\mathrm{P}(1,1,1)$ be perpendicular to the lines $\frac{x-4}{4}=\frac{y-1}{1}=\frac{z-1}{1}$ and $\frac{x-17}{1}=\frac{y-71}{1}=\frac{z}{0}$. Let the line L intersect the $y z-$ plane at the point Q. Another line parallel to L and passing through the point $\mathrm{S}(1,0,-1)$ intersects the $y z$-plane at the point R. Then the square of the area of the parallelogram PQRS is equal to $\_\_\_\_$.

The vertices B and C of a triangle ABC lie on the line $\frac{x}{1}=\frac{1-y}{-2}=\frac{\mathrm{z}-2}{3}$. The coordinates of A and B are $(1,6,3)$ and $(4,9, \alpha)$ respectively and C is at a distance of 10 units from B. The area (in sq. units) of $\triangle \mathrm{ABC}$ is :

Let $\mathrm{P}(\alpha, \beta, \gamma)$ be the point on the line $\frac{x-1}{2}=\frac{y+1}{-3}=z$ at a distance $4 \sqrt{14}$ from the point $(1,-1,0)$ and nearer to the origin. Then the shortest distance, between the lines $\frac{x-\alpha}{1}=\frac{y-\beta}{2}=\frac{z-\gamma}{3}$ and $\frac{x+5}{2}=\frac{y-10}{1}=\frac{z-3}{1}$, is equal to

If the distance of the point $(a, 2, 5)$ from the image of the point $(1, 2, 7)$ in the line $\dfrac{x}{1} = \dfrac{y-1}{1} = \dfrac{z-2}{2}$ is $4$, then the sum of all possible values of $a$ is equal to :

Let the vectors $\vec{a} = -\hat{i} + \hat{j} + 3\hat{k}$ and $\vec{b} = \hat{i} + 3\hat{j} + \hat{k}$. For some $\lambda, \mu \in \mathbb{R}$, let $\vec{c} = \lambda \vec{a} + \mu \vec{b}$. If $\vec{c} \cdot (3\hat{i} - 6\hat{j} + 2\hat{k}) = 10$ and $\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = -2$, then $|\vec{c}|^2$ is equal to:

Let $\vec{a_k} = (\tan\theta_k)\hat{i} + \hat{j}$ and $\vec{b_k} = \hat{i} - (\cot\theta_k)\hat{j}$, where $\theta_k = \dfrac{2^{k-1}\pi}{2^n + 1}$, for some $n \in \mathbb{N}$, $n > 5$. Then the value of $\dfrac{\displaystyle\sum_{k=1}^{n}|\vec{a_k}|^2}{\displaystyle\sum_{k=1}^{n}|\vec{b_k}|^2}$ is _____.

Let the point A be the foot of perpendicular drawn from the point $P(a, b, 0)$ on the line $\dfrac{x-1}{2} = \dfrac{y-2}{1} = \dfrac{z-\alpha}{3}$. If the midpoint of the line segment PA is $\left(0, \dfrac{3}{4}, \dfrac{-1}{4}\right)$, then the value of $a^2 + b^2 + \alpha^2$ is equal to:

Let a line $L_1$ pass through the origin and be perpendicular to the lines $L_2: \vec{r} = (3+t)\hat{i} + (2t-1)\hat{j} + (2t+4)\hat{k}$ and $L_3: \vec{r} = (3+2s)\hat{i} + (3+2s)\hat{j} + (2+s)\hat{k}$, $t, s \in \mathbb{R}$. If $(a, b, c)$, $a \in \mathbb{Z}$, is the point on $L_3$ at a distance of $\sqrt{17}$ from the point of intersection of $L_1$ and $L_2$, then $(a+b+c)^2$ is equal to ________.

Let $\overrightarrow{\mathrm{AB}}=2 \hat{i}+4 \hat{j}-5 \hat{k}$ and $\overrightarrow{\mathrm{AD}}=\hat{i}+2 \hat{j}+\lambda \hat{k}, \lambda \in \mathbb{R}$. Let the projection of the vector $\vec{v}=\hat{i}+\hat{j}+\hat{k}$ on the diagonal $\overrightarrow{\mathrm{AC}}$ of the parallelogram ABCD be of length one unit. If $\alpha, \beta$, where $\alpha>\beta$, be the roots of the equation $\lambda^{2} x^{2}-6 \lambda x+5=0$, then $2 \alpha-\beta$ is equal to

The square of the distance of the point $(-2, -8, 6)$ from the line $\dfrac{x-1}{1} = \dfrac{y-1}{2} = \dfrac{z}{-1}$ along the line $\dfrac{x+5}{1} = \dfrac{y+5}{-1} = \dfrac{z}{2}$ is equal to:

Let the image of the point $P(0, -5, 0)$ in the line $\dfrac{x-1}{2} = \dfrac{y}{1} = \dfrac{z+1}{-2}$ be the point $R$ and the image of the point $Q\left(0, \dfrac{-1}{2}, 0\right)$ in the line $\dfrac{x-1}{-1} = \dfrac{y+9}{4} = \dfrac{z+1}{1}$ be the point $S$. Then the square of the area of the parallelogram $PQRS$ is __________.

Let the line $\mathrm{L}_{1}$ be parallel to the vector $-3 \hat{i}+2 \hat{j}+4 \hat{k}$ and pass through the point (2,6,7), and the line $\mathrm{L}_{2}$ be parallel to the vector $2 \hat{i}+\hat{j}+3 \hat{k}$ and pass through the point $(4,3,5)$. If the line $\mathrm{L}_{3}$ is parallel to the vector $-3 \hat{i}+5 \hat{j}+16 \hat{k}$ and intersects the lines $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$ at the points C and D, respectively, then $|\overrightarrow{C D}|^{2}$ is equal to :

The square of the distance of the point of intersection of the lines $\vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(a\hat{i} - \hat{j})$, $a \neq 0$ and $\vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + a\hat{k})$ from the origin is:

If the image of the point $\mathrm{P}(a, 2, a)$ in the line $\frac{x}{2}=\frac{y+a}{1}=\frac{z}{1}$ is Q and the image of Q in the line $\frac{x-2 b}{2}=\frac{y-a}{1}=\frac{z+2 b}{-5}$ is P, then $a+b$ is equal to $\_\_\_\_$.

A line with direction ratios $1, -1, 2$ intersects the lines $\dfrac{x}{2} = \dfrac{y}{3} = \dfrac{z+1}{3}$ and $\dfrac{x+1}{-1} = \dfrac{y-2}{1} = \dfrac{z}{4}$ at the points $P$ and $Q$, respectively. If the length of the line segment $PQ$ is $\alpha$, then $225\alpha^2$ is equal to:

Let a line $L$ passing through the point $(1, 1, 1)$ be perpendicular to both the vectors $2\hat{i} + 2\hat{j} + \hat{k}$ and $\hat{i} + 2\hat{j} + 2\hat{k}$. If $P(a, b, c)$ is the foot of perpendicular from the origin on the line $L$, then the value of $34(a + b + c)$ is :

Let the foot of perpendicular from the point $(\lambda, 2, 3)$ on the line $\dfrac{x-4}{1} = \dfrac{y-9}{2} = \dfrac{z-5}{1}$ be the point $(1, \mu, 2)$. Then the distance between the lines $\dfrac{x-1}{2} = \dfrac{y-2}{3} = \dfrac{z+4}{6}$ and $\dfrac{x-\lambda}{2} = \dfrac{y-\mu}{3} = \dfrac{z+5}{6}$ is equal to:

Let L be the line $\frac{x+1}{2}=\frac{y+1}{3}=\frac{z+3}{6}$ and let S be the set of all points $(\mathrm{a}, \mathrm{b}, \mathrm{c})$ on L, whose distance from the line $\frac{x+1}{2}=\frac{y+1}{3}=\frac{z-9}{0}$ along the line $L$ is 7. Then $\sum_{(a, b, c) \in S}(a+b+c)$ is equal to :

Let $\vec{a}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}, \vec{b}=\hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}}$ and $\vec{c}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}}$. Let $\vec{v}$ be the vector in the plane of the vectors $\vec{a}$ and $\vec{b}$, such that the length of its projection on the vector $\vec{c}$ is $\frac{1}{\sqrt{14}}$. Then $|\vec{v}|$ is equal to

Let $P Q R$ be a triangle such that $\overrightarrow{P Q}=-2 \hat{i}-\hat{j}+2 \hat{k}$ and $\overrightarrow{\mathrm{PR}}=a \hat{\mathrm{i}}+b \hat{\mathrm{j}}-4 \hat{\mathrm{k}}, a, b \in \mathbb{Z}$. Let S be the point on QR, which is equidistant from the lines PQ and PR. If $|\overrightarrow{\mathrm{PR}}|=9$ and $\overrightarrow{\mathrm{PS}}=\hat{\mathrm{i}}-7 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$, then the value of $3 a-4 b$ is $\_\_\_\_$

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

The square of the distance of the point $P(5, 6, 7)$ from the line $\dfrac{x-2}{2} = \dfrac{y-5}{3} = \dfrac{z-2}{4}$ is equal to:

Let $\hat{u}$ and $\hat{v}$ be unit vectors inclined at an acute angle such that $|\hat{u}\times\hat{v}|=\dfrac{\sqrt{3}}{2}$. If $\vec{A}=\lambda\hat{u}+\hat{v}+(\hat{u}\times\hat{v})$, then $\lambda$ is equal to:

If $\vec{a}$ and $\vec{b}$ are two vectors such that $|\vec{a}| = 2$ and $|\vec{b}| = 3$, then the maximum value of $3\left|\left(3\vec{a} + 2\vec{b}\right)\right| + 4\left|\left(3\vec{a} - 2\vec{b}\right)\right|$ is :

Two adjacent sides of a parallelogram PQRS are given by $\vec{PQ} = \hat{j} + \hat{k}$ and $\vec{PS} = \hat{i} - \hat{j}$. If the side PS is rotated about the point P by an acute angle $\alpha$ in the plane of the parallelogram so that it becomes perpendicular to the side PQ, then $\sin^2\left(\dfrac{5\alpha}{2}\right) - \sin^2\left(\dfrac{\alpha}{2}\right)$ is equal to:

Let a vector $\overrightarrow{\mathrm{a}}=\sqrt{2} \hat{i}-\hat{j}+\lambda \hat{k}, \lambda>0$, make an obtuse angle with the vector $\overrightarrow{\mathrm{b}}=-\lambda^{2} \hat{i}+4 \sqrt{2} \hat{j}+4 \sqrt{2} \hat{k}$ and an angle $\theta, \frac{\pi}{6}<\theta<\frac{\pi}{2}$, with the positive $z$-axis. If the set of all possible values of $\lambda$ is $(\alpha, \beta)-\{\gamma\}$, then $\alpha+\beta+\gamma$ is equal to $\_\_\_\_$。

Let the image of the point $P(1, 6, a)$ in the line $L: \dfrac{x}{1} = \dfrac{y-1}{2} = \dfrac{z-a+1}{b}$, $b > 0$, be $\left(\dfrac{a}{3}, 0, a+c\right)$. If $S(\alpha, \beta, \gamma)$, $\alpha > 0$, is the point on $L$ such that the distance of $S$ from the foot of perpendicular from the point $P$ on $L$ is $2\sqrt{14}$, then $\alpha + \beta + \gamma$ is equal to:

Let the line $L$ pass through the point $(-3,5,2)$ and make equal angles with the positive coordinate axes. If the distance of L from the point $(-2, \mathrm{r}, 1)$ is $\sqrt{\frac{14}{3}}$, then the sum of all possible values of r is :

Let $(\alpha, \beta, \gamma)$ be the co-ordinates of the foot of the perpendicular drawn from the point $(5,4,2)$ on the line $\overrightarrow{\mathrm{r}}=(-\hat{i}+3 \hat{j}+\hat{k})+\lambda(2 \hat{i}+3 \hat{j}-\hat{k})$. Then the length of the projection of the vector $\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$ on the vector $6 \hat{i}+2 \hat{j}+3 \hat{k}$ is :

If $\left(2\alpha+1, \alpha^2-3\alpha, \dfrac{\alpha-1}{2}\right)$ is the image of $(\alpha, 2\alpha, 1)$ in the line $\dfrac{x-2}{3}=\dfrac{y-1}{2}=\dfrac{z}{1}$, then the possible value(s) of $\alpha$ is (are)