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

Let

P (α, β, γ)

be the point on the line

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

at a distance

414

from the point

(1, - 1, 0)

and nearer to the origin. Then the shortest distance, between the lines

\frac{x - α}{1} = \frac{y - β}{2} = \frac{z - γ}{3}

and

\frac{x + 5}{2} = \frac{y - 10}{1} = \frac{z - 3}{1}

, is equal to

Answer: (B)

Solution

Points on the first line are

P (2 λ + 1, - 3 λ - 1, λ)

, whose distance from

(1, - 1, 0)

14 λ^{2}

. Setting this to

414

gives

λ = \pm 4

; the point nearer the origin is

λ = - 4

, i.e.

P (- 7, 11, - 4)

. For the two lines through

(- 7, 11, - 4)

and

(- 5, 10, 3)

with directions

(1, 2, 3)

and

(2, 1, 1)

b_{1} \times b_{2} = (- 1, 5, - 3)

with

∣ b_{1} \times b_{2} ∣ = 35

, and the join of the points is

(2, - 1, 7)

. The shortest distance is

\frac{∣ ( 2 ) ( - 1 ) + ( - 1 ) ( 5 ) + ( 7 ) ( - 3 ) ∣}{35} = \frac{28}{35} = \frac{4 35}{5} = 4 \frac{7}{5} .

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