Q36 / 75 · JEE Main 2026 (22 January Shift 2) · Mathematics
Let L be the line $\frac{x+1}{2}=\frac{y+1}{3}=\frac{z+3}{6}$ and let S be the set of all points $(\mathrm{a}, \mathrm{b}, \mathrm{c})$ on L, whose distance from the line $\frac{x+1}{2}=\frac{y+1}{3}=\frac{z-9}{0}$ along the line $L$ is 7. Then $\sum_{(a, b, c) \in S}(a+b+c)$ is equal to :
- A6
- Correct answerB34
- C40
- D28
Solution
M is the point of intersection of $L_1$ & $L_2$
$\Rightarrow 2\lambda - 1 = 2\mu - 1, 3\lambda - 1 = 3\mu - 1, 6\lambda - 3 = 9$
$\Rightarrow \lambda = 2 = \mu$
$\Rightarrow M(3, 5, 9)$
Now let point P be $(2K - 1, 3K - 1, 6K - 3)$ on $L_2$
such that $PM = 7$
$\Rightarrow \sqrt{(2K - 4)^2 + (3K - 6)^2 + (6K - 12)^2} = 7$
$\Rightarrow 49K^2 + 196 - 196K = 49$
$\Rightarrow K^2 + 4 - 4K = 1$
$\Rightarrow K^2 - 4K + 3 = 0$
$\Rightarrow K = 1, 3$
So points P & Q are $(1, 2, 3)$ & $(5, 8, 15)$
So sum of all co-ordinates of P & Q $= 34$
$\Rightarrow 2\lambda - 1 = 2\mu - 1, 3\lambda - 1 = 3\mu - 1, 6\lambda - 3 = 9$
$\Rightarrow \lambda = 2 = \mu$
$\Rightarrow M(3, 5, 9)$
Now let point P be $(2K - 1, 3K - 1, 6K - 3)$ on $L_2$
such that $PM = 7$
$\Rightarrow \sqrt{(2K - 4)^2 + (3K - 6)^2 + (6K - 12)^2} = 7$
$\Rightarrow 49K^2 + 196 - 196K = 49$
$\Rightarrow K^2 + 4 - 4K = 1$
$\Rightarrow K^2 - 4K + 3 = 0$
$\Rightarrow K = 1, 3$
So points P & Q are $(1, 2, 3)$ & $(5, 8, 15)$
So sum of all co-ordinates of P & Q $= 34$
Hence, the correct answer is option: 'B'.