JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Three Dimensional Geometry

The square of the distance of the point (-2,-8,6) from the line

\frac{x - 1}{1} = \frac{y - 1}{2} = \frac{z}{- 1}

along the line

\frac{x + 5}{1} = \frac{y + 5}{- 1} = \frac{z}{2}

is equal to:

(A)3
(B)6✓ Correct
(C)8
(D)12

Answer: (B)

Solution

The measuring line through

A (- 2, - 8, 6)

with direction

(1, - 1, 2)

(t - 2, - t - 8, 2 t + 6)

. Setting this equal to a point

(s + 1, 2 s + 1, - s)

of the target line

\frac{x - 1}{1} = \frac{y - 1}{2} = \frac{z}{- 1}

gives

t - 2 = s + 1

- t - 8 = 2 s + 1

2 t + 6 = - s

, solved by

t = - 1

s = - 4

. The point of meeting is

B = (- 3, - 7, 4)

, and the squared distance is

A B^{2} = (- 1)^{2} + 1^{2} + (- 2)^{2} = 6

JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Single correct option

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