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
    $-\frac{3}{2}$
  • C
    $3$
  • D
    $-3$

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