Let the foot of perpendicular from the point $(\lambda, 2,3)$ on the line $\frac{x-4}{1}=\frac{y-9}{2}=\frac{z-5}{1}$ be the point $(1, \mu, 2)$. Then the distance between the lines $\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-2}{3}=\frac{\mathrm{z}+4}{6}$ and $\frac{\mathrm{x}-\lambda}{2}=\frac{\mathrm{y}-\mu}{3}=\frac{\mathrm{z}+5}{6}$ is equal to

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Three Dimensional Geometry

Let the foot of perpendicular from the point

(λ, 2, 3)

on the line

\frac{x - 4}{1} = \frac{y - 9}{2} = \frac{z - 5}{1}

be the point

(1, μ, 2)

. Then the distance between the lines

\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 4}{6}

and

\frac{x - λ}{2} = \frac{y - μ}{3} = \frac{z + 5}{6}

is equal to

Answer: (C)

Since

(1, μ, 2)

lies on the line,

\frac{1 - 4}{1} = \frac{μ - 9}{2} = \frac{2 - 5}{1} = - 3

, giving

μ = 3

. The segment from

(λ, 2, 3)

(1, 3, 2)

has direction

(1 - λ, 1, - 1)

; being perpendicular to the line's direction

(1, 2, 1)

(1 - λ) + 2 - 1 = 0 ⟹ λ = 2.

The two required lines both have direction

(2, 3, 6)

, so they are parallel, through

(1, 2, - 4)

and

(2, 3, - 5)

. With joining vector

(1, 1, - 1)

(1, 1, - 1) \times (2, 3, 6) = (9, - 8, 1), d = \frac{81 + 64 + 1}{4 + 9 + 36} = \frac{146}{7} .

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Single correct option