If the image of the point $\mathrm{P}(1,2, a)$ in the line $\frac{\mathrm{x}-6}{3}=\frac{\mathrm{y}-7}{2}=\frac{7-\mathrm{z}}{2}$ is $\mathrm{Q}(5, \mathrm{~b}, \mathrm{c})$, then $\mathrm{a}^{2}+\mathrm{b}^{2}+ \mathrm{c}^{2}$ is equal to

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Three Dimensional Geometry

If the image of the point

P (1, 2, a)

in the line

\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{7 - z}{2}

Q (5, b, c)

, then

a^{2} + b^{2} + c^{2}

is equal to

(A)293
(B)264
(C)298✓ Correct
(D)283

Answer: (C)

Solution

The line can be written as

(6 + 3 t, 7 + 2 t, 7 - 2 t)

with direction

(3, 2, - 2)

. The midpoint

M

P Q

(3, \frac{b + 2}{2}, \frac{a + c}{2})

and lies on the line; matching the

x

-coordinate,

3 = 6 + 3 t \Rightarrow t = - 1

, so

M = (3, 5, 9)

, giving

b = 8

and

a + c = 18

. Also

P Q = (4, 6, c - a)

is perpendicular to

(3, 2, - 2)

12 + 12 - 2 (c - a) = 0 \Rightarrow c - a = 12

. Solving,

c = 15

and

a = 3

, so

a^{2} + b^{2} + c^{2} = 9 + 64 + 225 = 298

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Single correct option

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