Q33 / 75 · JEE Main 2026 (21 January Shift 2) · Mathematics
Let the line $L$ pass through the point $(-3,5,2)$ and make equal angles with the positive coordinate axes. If the distance of L from the point $(-2, \mathrm{r}, 1)$ is $\sqrt{\frac{14}{3}}$, then the sum of all possible values of r is :
- Correct answerA10
- B6
- C12
- D16
Solution
Line $L$ through $(-3, 5, 2)$ making equal angles with axes has direction $(1, 1, 1)$.
Point $Q = (-2, r, 1)$, $\vec{PQ} = (1, r-5, -1)$.
$\vec{PQ} \times \vec{d} = (r-4, -2, 6-r)$.
$|\vec{PQ} \times \vec{d}|^2 = (r-4)^2 + 4 + (6-r)^2 = 2r^2 - 20r + 56$.
Distance $= \sqrt{\frac{2r^2 - 20r + 56}{3}} = \sqrt{\frac{14}{3}}$.
$2r^2 - 20r + 56 = 14 \Rightarrow r^2 - 10r + 21 = 0 \Rightarrow r = 3$ or $r = 7$.
Sum $= 3 + 7 = 10$.
Point $Q = (-2, r, 1)$, $\vec{PQ} = (1, r-5, -1)$.
$\vec{PQ} \times \vec{d} = (r-4, -2, 6-r)$.
$|\vec{PQ} \times \vec{d}|^2 = (r-4)^2 + 4 + (6-r)^2 = 2r^2 - 20r + 56$.
Distance $= \sqrt{\frac{2r^2 - 20r + 56}{3}} = \sqrt{\frac{14}{3}}$.
$2r^2 - 20r + 56 = 14 \Rightarrow r^2 - 10r + 21 = 0 \Rightarrow r = 3$ or $r = 7$.
Sum $= 3 + 7 = 10$.
Hence, the correct answer is option: 'A'.