If a line passing through (2, 1, 4) cuts off an intercept of minimum length between two non coplanar lines $x-6=\frac{y}{\alpha }=-z$ 

 and $x=\frac{y}{0}=z$   then $\alpha$  is

Line passes through $P\left(2,1,4\right)$  lie along shortest distance between two non – coplanar lines 

$L_1=\frac{x-6}{1}=\frac{y}{\alpha }=\frac{z}{-1}=k_1\Rightarrow A\left(6+k_1,\alpha k_1,-k_1\right)$

$L_2=\frac{x}{1}=\frac{y}{0}=\frac{z}{1}=k_2\Rightarrow B\left(k_2,0,k_2\right)$

$\begin{array}{l} \\ {\text{PA perpendicular to L}}_1\Rightarrow \left(6+k_1-2\right)1+\left(\alpha k_1-1\right)\alpha +\left(-k_1-4\right)\left(-1\right)=0\\ \\ {\alpha }^2{k}_1-\alpha +2k_1+8=0\\ PB{\text{\hspace{0.17em}is perpendicular to L}}_2\Rightarrow k_2=3\\ AB\text{\hspace{0.17em}perpendicualr to }{L}_1\\ \left(6+k_1-3\right)+\alpha \left(\alpha k_1\right)+\left(-1\right)\left(-k_1-3\right)=0\\ {\alpha }^2{k}_1+2k_1+6=0\\ \text{Hence, }\alpha \text{=2}\end{array}$