The shortest distance between the lines $x+1=2y=-12z$ $\text{ and }x=y+2=6z-6$ is 

$\frac{x-(-1)}{1}=\frac{y-0}{1/2}=\frac{z-0}{-1/12}\text{ i.e }\frac{x-(-1)}{-12}=\frac{y}{-6}=\frac{z}{1}$
Also $x=y+2=6(z-1)\Rightarrow \frac{x}{6}=\frac{y+2}{6}=\frac{z-1}{1}$

$\left({\mathrm{x}}_1,{\mathrm{y}}_1,{\mathrm{z}}_1\right)=(-1,0,0),\left({\mathrm{x}}_2,{\mathrm{y}}_2,{\mathrm{z}}_2\right)=(0,-2,1)$

$\left({\mathrm{b}}_1,{\mathrm{b}}_2,{\mathrm{b}}_3\right)=(-12,-6,1),\left({\mathrm{d}}_1,{\mathrm{d}}_2\cdot {\mathrm{d}}_3\right)=(6,6,1)$

$\mathrm{S}.\mathrm{D}=\frac{\left|\left[\begin{array}{ccc}\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{c}}& \overrightarrow{\mathrm{b}}& \overrightarrow{\mathrm{d}}\end{array}\right]\right|}{|\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{d}}|}$

$\left[\begin{array}{ccc}\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{c}}& \overrightarrow{\mathrm{b}}& \overrightarrow{\mathrm{d}}\end{array}\right]=\left|\begin{array}{ccc}-1& 2& -1\\ -12& -6& 1\\ 6& 6& 1\end{array}\right|$

$=-(-6,-6)-2(-12-6)-1(-72+36)=12+36+36=84$

$\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{d}}=\left|\begin{array}{ccc}\widehat{\mathrm{i}}& \widehat{\mathrm{j}}& \widehat{\mathrm{k}}\\ -12& -6& 1\\ 6& 6& 1\end{array}\right|$

$\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{d}}=\overrightarrow{\mathrm{i}}(-6-6)-\widehat{\mathrm{j}}(-12-6)+\widehat{\mathrm{k}}(-72+36)$

$\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{d}}=-12\widehat{\mathrm{i}}+18\widehat{\mathrm{j}}-36\widehat{\mathrm{k}}=-6(2\widehat{\mathrm{i}}-3\widehat{\mathrm{j}}+6\widehat{\mathrm{k}})$

$|\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{d}}|=6\sqrt{4+9+36}=6\times 7$

$\begin{array}{l}\text{ S.D }=\frac{84}{6\times 7}=2\end{array}$