The shortest distance between the lines $\frac{x-1}{0}=\frac{y+1}{-1}=\frac{z}{1}andx+y+z+1=0,2x-y+z+3=0is:$

$\begin{array}{l}{\mathrm{L}}_1:\frac{\mathrm{x}-1}{0}=\frac{\mathrm{y}+1}{-1}=\frac{\mathrm{z}}{1}\\ {\mathrm{L}}_2:\mathrm{x}+\mathrm{y}+\mathrm{z}+1=0,2\mathrm{x}-\mathrm{y}+\mathrm{z}+3=0\\ \text{ Equation of plane }(\pi =0)\text{ through }{\mathrm{L}}_2\text{ and parallel to }{\mathrm{L}}_1\text{ is }\\ \mathrm{x}+\mathrm{y}+\mathrm{z}+1+\lambda (2\mathrm{x}-\mathrm{y}+\mathrm{z}+3)=0\\ 0(1+2\lambda )-1(1-\lambda )+1(1+\lambda )=0\\ -1+\lambda +1+\lambda =0\\ \lambda =0\\ \mathrm{x}+\mathrm{y}+\mathrm{z}+1=0\end{array}$

$\text{ Shortest distance }=\text{ distance of }(1,-1,0)\text{ from }\mathrm{x}+\mathrm{y}+\mathrm{z}+1=0\text{ is }\frac{\left|p^{\text{'}}-\gamma +{}^2+1\right|}{\sqrt{1+1+1}}=\frac{1}{\sqrt{3}}$