Question 21 — JEE Main 22 Jan 2026 S Ii

Question

Let L be the line $\frac{\mathrm{x}+1}{2}=\frac{\mathrm{y}+1}{3}=\frac{\mathrm{z}+3}{6}$ and let S be the set of all points ( $\mathrm{a}, \mathrm{b}, \mathrm{c}$ ) on L , whose distance from the line $\frac{\mathrm{x}+1}{2}=\frac{\mathrm{y}+1}{3}=\frac{\mathrm{z}-9}{0}$ along the line L is 7 . Then $\sum_{(\mathrm{a}, \mathrm{b}, \mathrm{c}) \in \mathrm{S}}(\mathrm{a}+\mathrm{b}+\mathrm{c})$ is equal to
[JEE MAIN 22 Jan 2026 S II]
(1) 34
(2) 28
(3) 40
(4) 6

Solution

line $\mathrm{L} \equiv \frac{\mathrm{x}+1}{2}=\frac{\mathrm{y}+1}{3}=\frac{\mathrm{z}+3}{6}=\lambda=\mathrm{t}$
$\mathrm{A}(2 \lambda-1,3 \lambda-1,6 \lambda-3)$
$\mathrm{L}_{1}: \frac{\mathrm{x}+1}{2}=\frac{\mathrm{y}+1}{3}=\frac{\mathrm{z}-9}{0}=\mu$

$\mathrm{A}(2 \mu-1,3 \mu-1,9)$
$2 \lambda-1=2 \mu-1 \Rightarrow \lambda=\mu$
$6 \lambda-3=9 \Rightarrow 6 \lambda=12 \Rightarrow \lambda=2$
A(3,5,9)
Point $\mathrm{P}(2 \mathrm{t}-1,3 \mathrm{t}-1,6 \mathrm{t}-3)$
$\mathrm{AP}=7$
$\Rightarrow(2 \mathrm{t}-4)^{2}+(3 \mathrm{t}-6)^{2}+(6 \mathrm{t}-12)^{2}=49$
$\Rightarrow(\mathrm{t}-1)(\mathrm{t}-3)=0 \Rightarrow \mathrm{t}=1,3$
$\left.\begin{array}{ll}\mathrm{t}=1 & \mathrm{P}_{1}(1,2,3) \\ \mathrm{t}=3 & \mathrm{P}_{2}(5,8,15)\end{array}\right]$
$\sum_{\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{S}}(\mathrm{a}+\mathrm{b}+\mathrm{c})=1+2+3+5+8+15=34$ Answer $(1)$