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} $