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JEE MAIN 2025
04-04-2025 SHIFT-1
Question
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
Select the correct option:
A
48
B
42
C
40
D
46
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
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Mathematics
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