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3D Geometry question

2025 · 8 Apr · Shift 2 · Q26

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JEE Main
Mathematics
3D Geometry
2025 · 8 Apr · Shift 2 · Q26

3D Geometry question

2025 · 8 Apr · Shift 2 · Q26

JEE MainMathematics3D GeometryMCQ+4 / −1

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

A
$3$
B
$\frac{5\sqrt{2}}{3}$
C
$\frac{\sqrt{2}}{3}$
D
$4$

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