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.

Let's try to make it easier: if you take a minute to read carefully what is $\,W_2\,$ , you'll see that
$$\begin{pmatrix}x\\y\\z\end{pmatrix}\in W_2\iff y=0$$
Now, since an element of $\,w_1\in W_1\,$ has the general form
$$w_1=a\begin{pmatrix}1\\2\\3\end{pmatrix}+b\begin{pmatrix}2\\1\\1\end{pmatrix}=\begin{pmatrix}a+2b\\2a+b\\3a+b\end{pmatrix}$$
we see that $\,w_1\in W_2\iff 2a+b=0\,$
Since $\,\dim(W_1\cap W_2)=1\,$ (why?) , we get that the above condition completely determines the elements of the intersection, and thus
$$W_1\cap W_2=\left\{\,\begin{pmatrix}a+2b\\2a+b\\3a+b\end{pmatrix}\;;\;2a+b=0\,\right\}=\left\{\,\begin{pmatrix}-3a\\0\\a\end{pmatrix}\;;\;a\in\Bbb R\,\right\}=\operatorname{Span}\left\{\,\begin{pmatrix}-3\\0\\1\end{pmatrix}\,\right\}$$
