Given that, $4x + 4y -5z = 12 \quad ...(i) $
and $ 8x + 12y -13z = 32\quad ...(ii)$
we know direction ratio's of the line are
$(x, y, z) = (l, m, n)$
Eq. $(i)$ and $(ii)$ becomes
$ 4l + 4m -5n = 0\quad ...(iii)$
and $ 8l + 12m - 13n = 0 \quad ...(iv)$
or $\frac{l}{8} = \frac{m}{12} = \frac{n}{16} $
$\Rightarrow \frac{l}{2} = \frac{m}{3} = \frac{n}{4}$
Intersection point with $ z = 0$ is given by
$ 4x + 4y = 12\quad ...(v) $
$ \therefore 8x + 12y = 32 \quad ...(vi)$
On solving Eqs. $(v)$ and $(vi)$, we get $(1, 2, 0)$.
$\therefore $ Required line is $ \frac{x-1}{2} =\frac{y-2}{3} = \frac{z}{4}$.