Three Dimensional Geometry – Lines – JEE Main 23 Jan 2026 Shift 2

Solution:

Since Q is the image of P in Line 1 ($L_1$) and P is the image of Q in Line 2 ($L_2$), the midpoint of PQ must lie on both lines, and the vector $\vec{PQ}$ must be perpendicular to the direction vectors of both lines.

Let the direction vectors be:
$\vec{d_1} = 2\hat{i} + \hat{j} + \hat{k}$
$\vec{d_2} = 2\hat{i} + \hat{j} – 5\hat{k}$

The direction of $\vec{PQ}$ is parallel to $\vec{d_1} \times \vec{d_2}$:
$$ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 1 \\ 2 & 1 & -5 \end{vmatrix} = -6\hat{i} + 12\hat{j} $$
So, $\vec{PQ}$ is parallel to $-\hat{i} + 2\hat{j}$. The z-component is 0.

Let $P = (a, 2, a)$. Let $Q = (x, y, z)$.
$$ \vec{PQ} = (x-a, y-2, z-a) = k(-1, 2, 0) $$
This implies $z – a = 0 \Rightarrow z = a$.

Let M be the midpoint of PQ. $M = \left(\frac{x+a}{2}, \frac{y+2}{2}, a\right)$.
M lies on $L_1$:
$$ \frac{\frac{x+a}{2}}{2} = \frac{\frac{y+2}{2} + a}{1} = \frac{a}{1} $$
From z-coordinate term: $a/1 = a$. Consistent.
From x-coordinate term: $\frac{x+a}{4} = a \Rightarrow x = 3a$.
From y-coordinate term: $\frac{y+2+2a}{2} = a \Rightarrow y = -2$.
So, $Q = (3a, -2, a)$ and midpoint $M = (2a, 0, a)$.

Now, M also lies on $L_2$:
$$ \frac{2a – 2b}{2} = \frac{0 – a}{1} = \frac{a + 2b}{-5} $$
From first two parts: $a – b = -a \Rightarrow b = 2a$.
From last two parts: $-a = \frac{a + 4a}{-5} = -a$. (Consistent).

Now use orthogonality condition: $\vec{PQ} \perp \vec{d_2}$.
$\vec{PQ} = Q – P = (2a, -4, 0)$.
$\vec{PQ} \cdot \vec{d_2} = 0 \Rightarrow 2a(2) + (-4)(1) + 0(-5) = 0$.
$$ 4a – 4 = 0 \Rightarrow a = 1 $$

Then $b = 2a = 2$.
Value of $a + b = 1 + 2 = 3$.

Ans. (3)