Let $W_{1},W_{2}$ be sub-spaces of $\mathbb{R}^{4}$, find a subspace $W_{3}$ s.t $W_3\subset W_{2}$ and $W_{1}\oplus W_{3}=W_{1}+W_{2}$ Let $W_{1},W_{2}$ be  linear sub-spaces of $\mathbb{R}^{4}$.
$W_{1}=\text{sp}\{(1,2,3,4),(3,4,5,6),(7,8,9,10)\}$ 
$W_{2}=\text{sp }\{(x,y,z,w)| \ x+y=0\}$

Find a linear subspace of  $ \  \mathbb{R}^{4} \  ; W_{3}$ , such that: $W_{3}\subset W_{2} \  \text {*and*} \  W_{1}\oplus W_{3}=W_{1}+W_{2}$

My attempt: 
I applied Gaussian elimination on the vectors of $W_1$, such that:
$$W_{1}=\left\{ \left(\begin{array}{c}
x\\
y\\
z\\
w
\end{array}\right)=a\left(\begin{array}{c}
1\\
0\\
-2\\
-1
\end{array}\right)+b\left(\begin{array}{c}
0\\
1\\
-2\\
-1
\end{array}\right)|a,b\in\mathbb{R}\right\} $$ 
At that point I got stuck. I'm not sure how to continue.

HINT
We should write as a span (check again you derivation by RREF)
$$W_{1}=\left\{ \left(\begin{array}{c}
x\\
y\\
z\\
w
\end{array}\right)=s\left(\begin{array}{c}
1\\
1\\
1\\
1
\end{array}\right)+t\left(\begin{array}{c}
0\\
1\\
2\\
3
\end{array}\right)\right\}$$
and we can  also easily find that
$$W_{2}=\left\{ \left(\begin{array}{c}
x\\
y\\
z\\
w
\end{array}\right)=r\left(\begin{array}{c}
1\\
-1\\
0\\
0
\end{array}\right)+s\left(\begin{array}{c}
0\\
0\\
1\\
0
\end{array}\right)+t\left(\begin{array}{c}
0\\
0\\
0\\
1
\end{array}\right)\right\}$$
then check by RREF on the $5$ basis vectors that $W_{1}+W_{2}$ has dimension $4$ and finally select 2 basis vectors for $W_3$ from the basis vectors of $W_2$ such that  $W_{1}\oplus W_{3}=\mathbb{R^4}$.
