The equation of line of intersection of the planes $4x+ 4y-5z = 12, 8x+ 12y-13z = 32$can be written as: 

Let equation of line $\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}\dots \left(i\right)$

We have $4x+4y-5z=12 ...\left(ii\right)$

and  $8x+12y-13z=32 ...\left(iii\right)$

Let $z = 0.$Now putting$z = 0$ in (ii) and (iii), 

we get, $4x+4y=12,8x+12y=32,$ on solving these

equations, we get $x = 1, y = 2 .$

Equation of line passing through $(1,2,0)$ is 

$\frac{x-1}{l}=\frac{y-2}{m}=\frac{z-0}{n}$

From equation (i) and (ii), $4l+4m-5n=0$

and $8l+12m-13n=0$

$\Rightarrow \frac{l}{8}=\frac{m}{12}=\frac{n}{16}\text{ i.e. }\frac{l}{2}=\frac{m}{3}=\frac{n}{4}.$

Hence, equation of line is $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z}{4}.$