Question 24 — JEE Main 23 Jan 2026 S Ii

Solution

$\mathrm{L}_{1}: \frac{\mathrm{x}}{2}=\frac{\mathrm{y}+\mathrm{a}}{1}=\frac{\mathrm{z}}{1}=\lambda$
$\mathrm{A}(2 \lambda, \lambda-\mathrm{a}, \lambda)$
$\mathrm{L}_{2}: \frac{\mathrm{x}-2 \mathrm{~b}}{2}=\frac{\mathrm{y}-\mathrm{a}}{2}=\frac{\mathrm{z}+2 \mathrm{~b}}{-5}=\mu$
$\mathrm{A}(2 \mu+2 \mathrm{~b}, \mu+\mathrm{a},-5 \mu-2 \mathrm{~b})$
$\Rightarrow \lambda=\mu+\mathrm{b}, \lambda=\mu+2 \mathrm{a}, \mu+2 \mathrm{a}=-5 \mu-2 \mathrm{~b}$
From first two, $\mathrm{b}=2 \mathrm{a}$
From third, $6 \mu=-2 b-2 a=-4 a-2 a$

$\lambda=\mathrm{a}, \mu=-\mathrm{a}$
We have Point $\mathrm{A}(2 \mathrm{a}, 0, \mathrm{a})$ and $\mathrm{P}(\mathrm{a}, 2, \mathrm{a})$
$\mathrm{AP} \xrightarrow{\text { drs }} \mathrm{a},-2,0 ; \mathrm{L}_{1} \xrightarrow{\text { drs }} 2,1,1$
$\mathrm{AP} \perp \mathrm{L}_{1}$
$2 \mathrm{a}-2+0=0 \Rightarrow \mathrm{a}=1$
$\mathrm{b}=2 \mathrm{a}=2 \therefore \mathrm{a}+\mathrm{b}=3$ Answer(3)