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3D Geometry question

2025 · 28 Jan · Shift 2 · Q30

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JEE Main
Mathematics
3D Geometry
2025 · 28 Jan · Shift 2 · Q30

3D Geometry question

2025 · 28 Jan · Shift 2 · Q30

JEE MainMathematics3D GeometryMCQ+4 / −1

The square of the distance of the point

\left( \frac{15}{7}, \frac{32}{7}, 7 \right)

from the line

\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}

in the direction of the vector

\hat{i} + 4\hat{j} + 7\hat{k}

is:

A
66
B
54
C
41
D
44

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