The square of the distance of the point $\mathrm{P}(5,6,7)$ from the line $\frac{x-2}{2}=\frac{y-5}{3}=\frac{z-2}{4}$ is equal to :

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Three Dimensional Geometry

The square of the distance of the point

P (5, 6, 7)

from the line

\frac{x - 2}{2} = \frac{y - 5}{3} = \frac{z - 2}{4}

is equal to :

(A)3
(B)5
(C)6✓ Correct
(D)8

Answer: (C)

Solution

A general point on the line is

Q = (2 + 2 λ, 5 + 3 λ, 2 + 4 λ)

, and the line has direction

(2, 3, 4)

. With

P Q = (2 λ - 3, 3 λ - 1, 4 λ - 5)

, the perpendicularity condition

P Q \cdot (2, 3, 4) = 0

gives

4 λ - 6 + 9 λ - 3 + 16 λ - 20 = 0

, i.e.

29 λ = 29

, so

λ = 1

. The foot of the perpendicular is

Q = (4, 8, 6)

, and

P Q^{2} = (5 - 4)^{2} + (6 - 8)^{2} + (7 - 6)^{2} = 1 + 4 + 1

. The square of the distance of

P

from the line is

6

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Single correct option

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