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3D Geometry question

2025 · 4 Apr · Shift 1 · Q27

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JEE Main
Mathematics
3D Geometry
2025 · 4 Apr · Shift 1 · Q27

3D Geometry question

2025 · 4 Apr · Shift 1 · Q27

JEE MainMathematics3D GeometryMCQ+4 / −1

Let the shortest distance between the lines

\frac{x-3}{3}=\frac{y-\alpha}{-1}=\frac{z-3}{1}

and

\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-\beta}{4}

be

3 \sqrt{30}

. Then the positive value of

5 \alpha+\beta

is

A
42
B
40
C
48
D
46

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