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3D Geometry question

2024 · 4 Apr · Shift 2 · Q35

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JEE Main
Mathematics
3D Geometry
2024 · 4 Apr · Shift 2 · Q35

3D Geometry question

2024 · 4 Apr · Shift 2 · Q35

JEE MainMathematics3D GeometryMCQ+4 / −1

Let

\mathrm{P}

be the point of intersection of the lines

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

and

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

. Then, the shortest distance of

\mathrm{P}

from the line

4 x=2 y=z

is

A
$\frac{3 \sqrt{14}}{7}$
B
$\frac{5 \sqrt{14}}{7}$
C
$\frac{\sqrt{14}}{7}$
D
$\frac{6 \sqrt{14}}{7}$

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