JEE Main 2025 (8 Apr, Shift 2) Maths Q15: Skew Lines

Question

Let the values of

\lambda

for which the shortest distance between the lines

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

and

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

is

\frac{1}{\sqrt{6}}

be

\lambda_1

and

\lambda_2

. Then the radius of the circle passing through the points

(0, 0)

,

(\lambda_1, \lambda_2)

and

(\lambda_2, \lambda_1)

is

Official answer

Option 4

NTA final key.

Accepted Answer

Option 4: $\frac{5\sqrt{2}}{3}$

Answer

4

Answer

$\frac{\sqrt{2}}{3}$

Answer

3

Answer

$\frac{5\sqrt{2}}{3}$

JEE Main 8 April 2025, Shift 2, Maths Q15