JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Three Dimensional Geometry

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

(a, b, 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 :

(A)34✓ Correct
(B)28
(C)40
(D)6

Answer: (A)

Solution

Points of

L

are

(2 k - 1, 3 k - 1, 6 k - 3)

. The second line forces the common point by matching

z

6 k - 3 = 9

gives

k = 2

, so the lines meet at

M (3, 5, 9)

. The direction of

L

(2, 3, 6)

with magnitude

4 + 9 + 36 = 7

, so a point of

L

at distance

7

from

M

along

L

corresponds to shifting

k

\pm 1

, i.e.

k = 1

k = 3

(1, 2, 3) and (5, 8, 15) .

Therefore

(a, b, c) \in S \sum (a + b + c) = (1 + 2 + 3) + (5 + 8 + 15) = 34.

JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Single correct option

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