JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Three Dimensional Geometry

Let the line

L_{1}

be parallel to the vector

- 3 \hat{i} + 2 \hat{j} + 4 \hat{k}

and pass through the point

(2, 6, 7)

and the line

L_{2}

be parallel to the vector

2 \hat{i} + \hat{j} + 3 \hat{k}

and pass through the point

(4, 3, 5)

. If the line

L_{3}

is parallel to the vector

- 3 \hat{i} + 5 \hat{j} + 16 \hat{k}

and intersects the lines

L_{1}

and

L_{2}

at the points C and D, respectively, then

∣ C D ∣^{2}

is equal to :

Answer: (B)

Solution

Line

L_{1}

\frac{x - 2}{- 3} = \frac{y - 6}{2} = \frac{z - 7}{4} = λ_{1}

. Point C on

L_{1}

(- 3 λ_{1} + 2, 2 λ_{1} + 6, 4 λ_{1} + 7)

. Line

L_{2}

\frac{x - 4}{2} = \frac{y - 3}{1} = \frac{z - 5}{3} = λ_{2}

. Point D on

L_{2}

(2 λ_{2} + 4, λ_{2} + 3, 3 λ_{2} + 5)

. The direction ratios of

L_{3}

are

(- 3, 5, 16)

. For C and D to lie on

L_{3}

, the vector

C D

must be parallel to

(- 3, 5, 16)

. Thus

\frac{2 λ _{2} + 3 λ _{1} + 2}{- 3} = \frac{λ _{2} - 2 λ _{1} - 3}{5} = \frac{3 λ _{2} - 4 λ _{1} - 2}{16}

. Solving these equations gives

λ_{1} = - 3

and

λ_{2} = 2

. Then

C (11, 0, - 5)

and

D (8, 5, 11)

. Thus

∣ C D ∣^{2} = (8 - 11)^{2} + (5 - 0)^{2} + (11 - (- 5))^{2} = 9 + 25 + 256 = 290

JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Single correct option