For the lines $\frac{x}{2}=\frac{y}{-3}=\frac{z}{1}$ and $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$

Let $P\left(2r_1,-3r_1,r_1\right)$ and $\mathrm{Q}\left(3{\mathrm{r}}_2+2,-5{\mathrm{r}}_2+1,2{\mathrm{r}}_2-2\right)$ be any two points on lines, So that PQ line of shortest distance.

$2r_1-3r_2-2,-3r_1+5r_2-1,r_1-2r_2+2$

It is $\perp$ to both lines and solving gives

$r_1=\frac{31}{3},r_2=\frac{19}{3}$

$P=\left(\frac{62}{3},-31,\frac{31}{3}\right)Q=\left(21,\frac{-92}{3},\frac{32}{3}\right),\text{ drs of }PQ$

$\mathrm{as}\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)$