The shortest distance between the lines $\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{-1}$ and $\frac{x+3}{2}=\frac{y-6}{1}=\frac{z-5}{3}$ is

Given equation of lines$\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{-1}$ & 

$\frac{x+3}{2}=\frac{y-6}{1}=\frac{z-5}{3}$

Now let the point be $A=\left(3,2,1\right)$ on first line

and $B=\left(-3,6,5\right)$ on second line, so $\overrightarrow{BA}=6\hat{\mathrm{i}}-4\hat{\mathrm{j}}-4\hat{k}$

And direction ratio's be $\overrightarrow{n_1}=2\hat{i}+3\hat{j}-\hat{k}$ &

$\overrightarrow{n_2}=2\hat{i}+\hat{i}-3\hat{k}$

$\overrightarrow{n_1}\times \overrightarrow{n_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& -1\\ 2& 1& 3\end{array}\right|$$=10\hat{i}-8\hat{j}-4\hat{k}$

We know that shortest Distance $=\frac{\left[\begin{array}{ccc}\overrightarrow{BA}& \overrightarrow{n_1}& \overrightarrow{n_2}\end{array}\right]}{\left|\overrightarrow{n_1}\times \overrightarrow{n_2}\right|}$

$$$\left[\begin{array}{lll}\overrightarrow{BA}& \overrightarrow{n_1}& \overrightarrow{n_2}\end{array}\right]=\left(6\hat{\mathrm{i}}-4\hat{\mathrm{j}}-4\hat{k}\right).\left(10\hat{i}-8\hat{j}-4\hat{k}\right)=60+32+16=108$

$\left|\overrightarrow{n_1}\times \overrightarrow{n_2}\right|=\sqrt{100+64+16}=\sqrt{180}$

$S.D=\frac{108}{\sqrt{180}}=\frac{108}{6\sqrt{5}}=\frac{18}{\sqrt{5}}$