Question 20 — JEE Main 22 Jan 2026 (S I)

Solution

line $\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$
$\mathrm{PQ} \xrightarrow{\text { drs }} 5-1, \mathrm{~b}-2, \mathrm{c}-\mathrm{a}$
or $4, \mathrm{~b}-2, \mathrm{c}-\mathrm{a}$
line $\xrightarrow{\text { drs }} 3,2,-2$
$P Q \perp$ line $\Rightarrow 4(3)+2(b-2)-2(c-a)=0$
$\Rightarrow \mathrm{b}-\mathrm{c}+\mathrm{a}+4=0$

$\mathrm{M}=$ Mid Point of PQ i.e. $\mathrm{P}(1,2, \mathrm{a}) \mathrm{Q}(5, \mathrm{~b}, \mathrm{c})$,
$\mathrm{M}\left(3, \frac{\mathrm{~b}+2}{2}, \frac{\mathrm{a}+\mathrm{c}}{2}\right)$ lies on given line
$\frac{3-6}{3}=\frac{\frac{\mathrm{b}+2}{2}-7}{2}=\frac{\frac{\mathrm{a}+\mathrm{c}}{2}-7}{-2}$
$\Rightarrow-1=\frac{\mathrm{b}-12}{4}=\frac{\mathrm{a}+\mathrm{c}-14}{-4}$
$\mathrm{b}=8 ; \mathrm{a}+\mathrm{c}-14=4 \Rightarrow \mathrm{a}+\mathrm{c}=18$
from (1): $8-c+a+4=0$
$\Rightarrow \mathrm{c}=\mathrm{a}+12$
From (2), $\mathrm{a}+\mathrm{a}+12=18 \Rightarrow \mathrm{a}=3$
$\mathrm{c}=3+12=15$
$\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}=9+64+225=298$ Answer $(3)$