$\text{ The equation of the line of intersection of planes }4x+4y-5z=12\text{ and }8x+12y-13z=32$ can be written as

$Given equation of planes are$

$\begin{array}{l}4x+4y-5z=12-----\left(\mathrm{i}\right)\\ \text{ And }8x+12y-13z=32-----\left(\mathrm{ii}\right)\end{array}$

$\text{ Let DR's of required line be }(l,m,n)$

From Eq’s (i) and (ii) we get 

$4l+4m-5n=0$

$\text{ And }8l+12m-13n=0$

$\begin{array}{r}\Rightarrow \frac{l}{8}=\frac{m}{12}=\frac{n}{16}\\ \Rightarrow \frac{l}{2}=\frac{m}{3}=\frac{n}{4}\end{array}$

$\text{ Now we take intersection point with }z=0\text{ is given by }$

$\begin{array}{l}4x+4y=12----\left(\mathrm{iii}\right)\\ \text{ And }8x+12y=32----\left(\mathrm{iv}\right)\end{array}$

$\begin{array}{l}\text{ On solving Equations (i) and (ii) we get the point }(1,2,0)\\ \text{ Required line is }\frac{x-1}{2}=\frac{y-2}{3}=\frac{z}{4}\end{array}$