If the shortest distance between the lines (x-1)/(α)=(y+1)/-1=z/1,(α ≠-1) and x+y+z+1=0=2 x-y+z+3 is 1/(√ 3), then a value of α is:

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