The Cartesian equation of the line passing through the point  $(-1,3,-2)$  and perpendicular to  the lines $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$  and $\frac{x+2}{-3}=\frac{y-1}{2}=\frac{z+1}{5}$   is 

Let (a, b, c) be the d.r’s of the required line 
This line is perpendicular to the lines  $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ ,   $\frac{x+2}{-3}=\frac{y-1}{2}=\frac{z+1}{5}$
 $\Rightarrow \frac{a}{10-6}=\frac{b}{-9-5}=\frac{c}{2+6}\Rightarrow \frac{a}{4}=\frac{b}{-14}=\frac{c}{8}\Rightarrow \frac{a}{2}=\frac{b}{-7}=\frac{c}{4}.$
The equation of the required line are $\frac{x+1}{2}=\frac{y-3}{-7}=\frac{z+2}{4}$