JEE Main 2026 — Mathematics Vectors & 3D Geometry

2026mcqhard

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:

Official previous-year question

Held on 6 Apr 2026 · Verified 6 Jul 2026.

Options

A
$19$
B
$20$
C
$21$
D
$22$

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Solution

Let the foot of the perpendicular from $P(1, 6, a)$ to the line $L$ be $M$ . Since the image of $P$ is $P'\left(\dfrac{a}{3}, 0, a+c\right)$ , $M$ is the midpoint of $P$ and $P'$ .

$M = \left(\dfrac{1 + \dfrac{a}{3}}{2}, \dfrac{6 + 0}{2}, \dfrac{a + a + c}{2}\right) = \left(\dfrac{a+3}{6}, 3, a + \dfrac{c}{2}\right)$

Since $M$ lies on the line $L: \dfrac{x}{1} = \dfrac{y-1}{2} = \dfrac{z-a+1}{b}$ , we substitute its coordinates into the equation of the line:

$\dfrac{\dfrac{a+3}{6}}{1} = \dfrac{3-1}{2} = \dfrac{a + \dfrac{c}{2} - a + 1}{b}$

$\dfrac{a+3}{6} = 1 = \dfrac{\dfrac{c}{2} + 1}{b}$

From $\dfrac{a+3}{6} = 1$ , we get $a = 3$ .

From $\dfrac{\dfrac{c}{2} + 1}{b} = 1$ , we get $2b = c + 2 \Rightarrow c = 2b - 2$ .

The vector $\overrightarrow{PP'}$ is perpendicular to the direction vector of the line $L$ , which is $\vec{v} = \hat{i} + 2\hat{j} + b\hat{k}$ .

$\overrightarrow{PP'} = \left(\dfrac{a}{3} - 1\right)\hat{i} + (0 - 6)\hat{j} + (a+c - a)\hat{k} = \left(\dfrac{3}{3} - 1\right)\hat{i} - 6\hat{j} + c\hat{k} = -6\hat{j} + c\hat{k}$

Taking the dot product $\overrightarrow{PP'} \cdot \vec{v} = 0$ :

$0(1) - 6(2) + c(b) = 0 \Rightarrow bc = 12$

Substituting $c = 2b - 2$ into $bc = 12$ :

$b(2b - 2) = 12 \Rightarrow 2b^2 - 2b - 12 = 0 \Rightarrow b^2 - b - 6 = 0$

$(b-3)(b+2) = 0$

Since $b > 0$ , we have $b = 3$ . Then $c = 2(3) - 2 = 4$ .

The coordinates of $M$ are $\left(\dfrac{3+3}{6}, 3, 3 + \dfrac{4}{2}\right) = (1, 3, 5)$ .

The equation of the line $L$ is $\dfrac{x}{1} = \dfrac{y-1}{2} = \dfrac{z-2}{3}$ . Any point $S$ on the line can be written as $(t, 2t+1, 3t+2)$ .

The distance $SM$ is given as $2\sqrt{14}$ .

$SM^2 = (t-1)^2 + (2t+1-3)^2 + (3t+2-5)^2 = 56$

$(t-1)^2 + 4(t-1)^2 + 9(t-1)^2 = 56$

$14(t-1)^2 = 56 \Rightarrow (t-1)^2 = 4 \Rightarrow t - 1 = \pm 2$

$t = 3$ or $t = -1$

Since $S(\alpha, \beta, \gamma) = (t, 2t+1, 3t+2)$ and $\alpha > 0$ , we must have $t = 3$ .

Thus, $S = (3, 7, 11)$ .

$\alpha + \beta + \gamma = 3 + 7 + 11 = 21$

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Solution

Let the foot of the perpendicular from $P(1, 6, a)$ to the line $L$ be $M$ . Since the image of $P$ is $P'\left(\dfrac{a}{3}, 0, a+c\right)$ , $M$ is the midpoint of $P$ and $P'$ .

$M = \left(\dfrac{1 + \dfrac{a}{3}}{2}, \dfrac{6 + 0}{2}, \dfrac{a + a + c}{2}\right) = \left(\dfrac{a+3}{6}, 3, a + \dfrac{c}{2}\right)$

Since $M$ lies on the line $L: \dfrac{x}{1} = \dfrac{y-1}{2} = \dfrac{z-a+1}{b}$ , we substitute its coordinates into the equation of the line:

$\dfrac{\dfrac{a+3}{6}}{1} = \dfrac{3-1}{2} = \dfrac{a + \dfrac{c}{2} - a + 1}{b}$

$\dfrac{a+3}{6} = 1 = \dfrac{\dfrac{c}{2} + 1}{b}$

From $\dfrac{a+3}{6} = 1$ , we get $a = 3$ .

From $\dfrac{\dfrac{c}{2} + 1}{b} = 1$ , we get $2b = c + 2 \Rightarrow c = 2b - 2$ .

The vector $\overrightarrow{PP'}$ is perpendicular to the direction vector of the line $L$ , which is $\vec{v} = \hat{i} + 2\hat{j} + b\hat{k}$ .

$\overrightarrow{PP'} = \left(\dfrac{a}{3} - 1\right)\hat{i} + (0 - 6)\hat{j} + (a+c - a)\hat{k} = \left(\dfrac{3}{3} - 1\right)\hat{i} - 6\hat{j} + c\hat{k} = -6\hat{j} + c\hat{k}$

Taking the dot product $\overrightarrow{PP'} \cdot \vec{v} = 0$ :

$0(1) - 6(2) + c(b) = 0 \Rightarrow bc = 12$

Substituting $c = 2b - 2$ into $bc = 12$ :

$b(2b - 2) = 12 \Rightarrow 2b^2 - 2b - 12 = 0 \Rightarrow b^2 - b - 6 = 0$

$(b-3)(b+2) = 0$

Since $b > 0$ , we have $b = 3$ . Then $c = 2(3) - 2 = 4$ .

The coordinates of $M$ are $\left(\dfrac{3+3}{6}, 3, 3 + \dfrac{4}{2}\right) = (1, 3, 5)$ .

The equation of the line $L$ is $\dfrac{x}{1} = \dfrac{y-1}{2} = \dfrac{z-2}{3}$ . Any point $S$ on the line can be written as $(t, 2t+1, 3t+2)$ .

The distance $SM$ is given as $2\sqrt{14}$ .

$SM^2 = (t-1)^2 + (2t+1-3)^2 + (3t+2-5)^2 = 56$

$(t-1)^2 + 4(t-1)^2 + 9(t-1)^2 = 56$

$14(t-1)^2 = 56 \Rightarrow (t-1)^2 = 4 \Rightarrow t - 1 = \pm 2$

$t = 3$ or $t = -1$

Since $S(\alpha, \beta, \gamma) = (t, 2t+1, 3t+2)$ and $\alpha > 0$ , we must have $t = 3$ .

Thus, $S = (3, 7, 11)$ .

$\alpha + \beta + \gamma = 3 + 7 + 11 = 21$

JEE Main 2026 — Mathematics Vectors & 3D Geometry

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JEE Main 2026 — Mathematics Vectors & 3D Geometry

Options

Did you get this right?

Solution

Your note

JEE Main Vectors & 3D Geometry in other years

More JEE Main Vectors & 3D Geometry questions

Other Mathematics topics for JEE Main

Algebra

Calculus

Coordinate Geometry

Probability & Statistics

Trigonometry

Keep practicing Vectors & 3D Geometry