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

$ \sqrt{2} = \frac{\left| \begin{matrix} -1 & 1 & 3 \ \alpha & -1 & -\alpha \ \alpha & 2 & 2\alpha \end{matrix} \right|}{\left| \begin{matrix} i & j & k \ \alpha & -1 & -\alpha \ \alpha & 2 & 2\alpha \end{matrix} \right|} $

$ \sqrt{2} = \frac{-1(-2\alpha + 2\alpha) - 1(2\alpha^2 + \alpha^2) + 3(2\alpha + \alpha)}{\sqrt{i(-2\alpha + 2\alpha)^2 + j(2\alpha^2 + \alpha^2)^2 + k(2\alpha + \alpha)^2}} $

$ \sqrt{2} = \frac{-3\alpha^2 + 9\alpha}{\sqrt{9\alpha^4 + 9\alpha^2}} $

$ \sqrt{2} = \frac{\alpha + 3}{\sqrt{\alpha^2 + 1}} $

$ \Rightarrow 2\alpha^2 + 2 = \alpha^2 + 9 - 6\alpha $

$ \alpha^2 + 6\alpha - 7 = 0 $

$ (\alpha + 7)(\alpha - 1) = 0 $

$ \alpha = -7,; 1 $

sum $ = -7 + 1 = -6 $