Solution
= Solution
The plane normals are $n_1=(2,-1,-1)$ and $n_2=(3,-1,-3)$. Their <cross product> is $(2,3,1)$, perpendicular to both normals and hence parallel to the intersection line. Setting $z=0$ in both equations gives the common point $(1,-1,0)$. Therefore
$$
\boxed{r=(1,-1,0)+t(2,3,1),\qquad t\in\mathbb R.}
$$
The normals are not parallel, so their intersection is a unique line.