3d Geometry

The sum of all values of $\alpha$, for which the shortest distance between the lines $\frac{x+1}{\alpha}=\frac{y-2}{-1}=\frac{z-4}{-\alpha}$ and $\frac{x}{\alpha}=\frac{y-1}{2}=\frac{z-1}{2 \alpha}$ is $\sqrt{2}$, is