JEE Math Practice Question
If the shortest distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x}{1}=\frac{y}{\alpha}=\frac{z-5}{1}$ is $\frac{5}{\sqrt{6}}$, then the sum of all possible values of $\alpha$ is
- A.$\frac{3}{2}$
- B.$3$
- C.$-3$
- D.$-\frac{3}{2}$
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