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 :

The given lines are 

$L_1:\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{-1}$

$$L_2:\frac{x+3}{2}=\frac{y-6}{1}=\frac{z-5}{3}$$

Vectors along the lines are $\overrightarrow{p}=2i+3j-k and \overrightarrow{q}=2i+j+3k$ and position vectors of the point on the lines are $\overrightarrow{a_1}=3i+2j+k$   and  $\overrightarrow{a_2}=-3i+6j+5k$

Now, $\overrightarrow{p}\times \overrightarrow{q}=\left|\begin{array}{ccc}i& j& k\\ 2& 3& -1\\ 2& 1& 3\end{array}\right|=10\hat{i}-8\hat{j}-4\hat{k}$ and ${\overrightarrow{a}}_2-{\overrightarrow{a}}_1=6\hat{i}-4\hat{j}-4\hat{k}$

The shortest distance between the above two lines is SD$SD=\frac{\left|\left(p\times q\right)·\left(a_2-a_1\right)\right|}{\left|\left(p\times q\right)\right|}$

Therefore, $SD=\left|\frac{60+32+16}{\sqrt{100+64+16}}\right|=\frac{108}{6\sqrt{5}}=\boxed{\frac{18}{\sqrt{5}}}$