$\text{ The point of intersection of the lines }\frac{x-5}{3}=\frac{y-7}{-1}=\frac{z+2}{1}$$\text{ and }\frac{x+3}{-36}=\frac{y-3}{2}=\frac{z-6}{4}\text{ is }$

 Given lines are

$\begin{array}{l}\frac{x-5}{3}=\frac{y-7}{-1}=\frac{z+2}{1}=r_1\\ \text{ and }\frac{x+3}{-36}=\frac{y-3}{2}=\frac{z-6}{4}=r_2\\ \therefore x=3r_1+5=-36r_2-3\\ y=-r_1+7=3+2r_2\end{array}$

$\text{ and }z=r_1-2=4r_2+6$

On solving, we get 

$x=21,y=\frac{5}{3},z=\frac{10}{3}$