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3D Geometry question

2024 · 6 Apr · Shift 1 · Q39

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JEE Main
Mathematics
3D Geometry
2024 · 6 Apr · Shift 1 · Q39

3D Geometry question

2024 · 6 Apr · Shift 1 · Q39

JEE MainMathematics3D GeometryMCQ+4 / −1

The shortest distance between the lines

\frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5}

and

\frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3}

is

A
$8 \sqrt{3}$
B
$6 \sqrt{3}$
C
$5 \sqrt{3}$
D
$4 \sqrt{3}$

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