3D Geometry – Lines – JEE Main 28 Jan 2026 Shift 2

Solution:

**Find the image $Q$ of $P(3,2,1)$ in the first line.

Let the line be $L_1: \frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1} = r$.
A general point on $L_1$ is $N(r+1, 2r, r+1)$.

The vector $\vec{PN} = (r+1-3)\hat{i} + (2r-2)\hat{j} + (r+1-1)\hat{k} = (r-2)\hat{i} + (2r-2)\hat{j} + (r)\hat{k}$.
Since $\vec{PN}$ is perpendicular to the line’s direction vector $\vec{d_1} = \hat{i} + 2\hat{j} + \hat{k}$:
$$1(r-2) + 2(2r-2) + 1(r) = 0$$
$$r – 2 + 4r – 4 + r = 0$$
$$6r = 6 \Rightarrow r = 1$$

The foot of the perpendicular is $N(2, 2, 2)$.
Since $N$ is the midpoint of $PQ$, let $Q = (x, y, z)$:
$$\frac{x+3}{2} = 2 \Rightarrow x = 1$$
$$\frac{y+2}{2} = 2 \Rightarrow y = 2$$
$$\frac{z+1}{2} = 2 \Rightarrow z = 3$$
So, $Q(1, 2, 3)$.

**Find distance of $Q(1, 2, 3)$ from the second line.

Line $L_2: \frac{x-9}{3}=\frac{y-9}{2}=\frac{z-5}{-2}$.
Point $A(9, 9, 5)$ lies on $L_2$. Direction vector $\vec{d_2} = 3\hat{i} + 2\hat{j} – 2\hat{k}$.

Vector $\vec{AQ} = (1-9)\hat{i} + (2-9)\hat{j} + (3-5)\hat{k} = -8\hat{i} – 7\hat{j} – 2\hat{k}$.
Magnitude $|\vec{AQ}|^2 = (-8)^2 + (-7)^2 + (-2)^2 = 64 + 49 + 4 = 117$.

Projection of $\vec{AQ}$ on $L_2$:
$$p = \frac{|\vec{AQ} \cdot \vec{d_2}|}{|\vec{d_2}|} = \frac{|(-8)(3) + (-7)(2) + (-2)(-2)|}{\sqrt{3^2+2^2+(-2)^2}}$$
$$p = \frac{|-24 – 14 + 4|}{\sqrt{9+4+4}} = \frac{34}{\sqrt{17}} = 2\sqrt{17}$$

Perpendicular distance $d$:
$$d^2 = |\vec{AQ}|^2 – p^2$$
$$d^2 = 117 – (2\sqrt{17})^2 = 117 – 4(17) = 117 – 68 = 49$$
$$d = \sqrt{49} = 7$$

Ans. (3)