Common basis for subspace intersection 

Let $ W_1 = \textrm{span}\left\{\begin{pmatrix}1\\2\\3\end{pmatrix}, \begin{pmatrix}2\\1\\1\end{pmatrix}\right\}$, and $ W_2 = \textrm{span}\left\{\begin{pmatrix}1\\0\\1\end{pmatrix}, \begin{pmatrix}3\\0\\-1\end{pmatrix}\right\}$. Find a basis for $W_1 \cap W_2$


I first thought of solving the augmented matrix:
$ \begin{pmatrix}1 && 2 && 1 && 3\\2 && 1 && 0 && 0\\3 && 1 && 1 && -1\end{pmatrix}$
But this matrix can have 3 pivots and so it's column space dimension can be at most 3 (which doesn't make sense since the basis I'm looking for must have dimension 2. 
So, what is the correct way to solve these exercise.

The two sub spaces are not the same, because $W_2$ has no extent in the second axis while $W_1$ does. The intersection is then a line in the $xz$ plane, which $W_2$ spans. If you can find a vector in $W_1$ in that plane that is your basis
