# If the shortest distance between the lines $\frac{x-1}{\alpha} = \frac{y+1}{-1} = \frac{z}{1},\; (\alpha \neq -1)$ and $x+y+z+1=0=2x-y+z+3$ is $\frac{1}{\sqrt{3}}$, then a value of $\alpha$ is :
( A ) $-\frac{19}{16}$
( B ) $\frac{19}{32}$
( C ) $-\frac{16}{19}$
( D ) $\frac{32}{19}$