Let the shortest distance between the lines (x-3)/3 = (y-α)/(-1)… | JEE Main 2025 Mathematics PYQ

MCQMediumJEE Main 2025 · 4 April, Shift 1Skew Lines & Shortest Distance

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
$46$
C
$48$
D
$40$

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