If the shortest distance between the lines $\large\frac{x-1}{\alpha } = \frac{y+1}{-1}= \frac{z}{1}$$, (\alpha \neq -1)$ and $x+y+z+1=0=2x-y+z+3$ is $\large\frac{1}{\sqrt 3}$, then a value of $\alpha$ is

Question

$\begin{array}{1 1} \frac{-16}{19} \\ \frac{19}{16} \\ \frac{32}{19} \\\frac{19}{32}\end{array}$