Let us change the line into symmetric form. 

$x+y+z+1=0=2x-y+z+3$

Put $z=1$ , so we get $x+y+2=0$ and $2x-y+4=\mathrm{0 }$

We will get  $x=-2$, $y=0$

$\therefore$ The point $\left(-2,0,1\right)$ lies on the line and perpendicular vector will come from  $\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 1\\ 2& -1& 1\end{array}\right|=2\hat{i}+\hat{j}-3\hat{k}$

So the equation of line would be $\frac{x+2}{2}=\frac{y}{1}=\frac{z-1}{-3}$
And the other line $\frac{x-1}{\alpha }=\frac{y+1}{-1}=\frac{z}{1}$
Shortest distance would be    $\mathrm{D}=\frac{\left|\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)·\left(\overrightarrow{b_1}\times \overrightarrow{b_2}\right)\right|}{\left|\left(\overrightarrow{b_1}\times \overrightarrow{b_2}\right)\right|}$

When   $\overrightarrow{a_1}=\left(-2\hat{i}+0\hat{j}+1\hat{k}\right)$

$\overrightarrow{a_2}=\left(\hat{i}-\hat{j}+0\hat{k}\right)$

$\overrightarrow{b_1}=2\hat{i}+\hat{j}-3\hat{k}$

$\overrightarrow{b_2}=\alpha \hat{i}-\hat{j}+\hat{k}$

$\mathrm{D}=$  $\left|\frac{\left|\begin{array}{ccc}3& -1& -1\\ 2& \mathrm{ }\mathrm{ }1& -3\\ \mathrm{\alpha }& -1& \mathrm{ }\mathrm{ }1\end{array}\right|}{\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& \mathrm{ }\mathrm{ }1& -3\\ \mathrm{\alpha }& -1& \mathrm{ }\mathrm{ }\mathrm{ }1\end{array}\right|}\right|$ 

$=\mathrm{ }\frac{\left|3\left(1-3\right)+1\left(2+3\mathrm{\alpha }\right)+1\left(2+\mathrm{\alpha }\right)\right|}{\left|-2\hat{i}-\hat{\mathit{j}}\left(2+3\mathrm{\alpha }\right)+\hat{\mathit{k}}\left(-2-\mathrm{\alpha }\right)\right|}$

$\Rightarrow$  $\left|\frac{-6+2+3\alpha +2+\alpha }{\sqrt{4+{(2+3\alpha )}^2+{(2+\alpha )}^2}}\right|=\frac{1}{\sqrt{3}}$

$\frac{\left|4\alpha -2\right|}{\sqrt{4+4+12\alpha +9{\alpha }^2+4+4\alpha +{\alpha }^2}}=\frac{1}{\sqrt{3}}$

$\left|\frac{4\alpha -2}{\sqrt{10{\alpha }^2+16\alpha +12}}\right|=\frac{1}{\sqrt{3}}$

$\Rightarrow \mathrm{2\alpha }\left(19\alpha -32\right)=0$

$\therefore \alpha =\frac{32}{19}$