3D Geometry – Line Intersection and Distance – JEE Main 22 Jan 2026 Shift 2

Solution:

First, we find the point of intersection $M$ of line $L$ and the second line $L_2$.
$$ L: \frac{x+1}{2} = \frac{y+1}{3} = \frac{z+3}{6} = \lambda $$
$$ L_2: \frac{x+1}{2} = \frac{y+1}{3} = \frac{z-9}{0} = \mu $$

Equating coordinates:
$$ 2\lambda – 1 = 2\mu – 1 \Rightarrow \lambda = \mu $$
$$ 3\lambda – 1 = 3\mu – 1 $$
$$ 6\lambda – 3 = 9 \Rightarrow 6\lambda = 12 \Rightarrow \lambda = 2 $$
So, $\lambda = 2$ and $\mu = 2$.
[cite_start]The point of intersection is $M(3, 5, 9)$[cite: 221, 222, 223].

Now, let $P$ be a point on line $L$ with parameter $K$:
$$ P(2K-1, 3K-1, 6K-3) $$
We are given that the distance $PM = 7$.
Using the distance formula between $P(2K-1, 3K-1, 6K-3)$ and $M(3, 5, 9)$:
$$ \sqrt{(2K-1-3)^2 + (3K-1-5)^2 + (6K-3-9)^2} = 7 $$
$$ \sqrt{(2K-4)^2 + (3K-6)^2 + (6K-12)^2} = 7 $$

Squaring both sides and expanding:
$$ (2K-4)^2 + (3K-6)^2 + (6K-12)^2 = 49 $$
$$ 4(K-2)^2 + 9(K-2)^2 + 36(K-2)^2 = 49 $$
$$ 49(K-2)^2 = 49 $$
$$ (K-2)^2 = 1 $$

$$ K – 2 = \pm 1 $$
$$ K = 3 \quad \text{or} \quad K = 1 $$

Substituting values of $K$ to find points $P$ and $Q$:
For $K=1$: $P(1, 2, 3)$
For $K=3$: $Q(5, 8, 15)$

Sum of all coordinates of $P$ and $Q$:
$$ \text{Sum} = (1+2+3) + (5+8+15) $$
$$ \text{Sum} = 6 + 28 = 34 $$

Ans. (1)