RREF Practice Check I am going over the following exercise

Find a condition on $a,b,c$ so that $(a,b,c)$ in $\mathbb{R}^{3}$
  belongs to the space spanned by $u = (2,1,0)$, $v=(1,-1,2)$, and $w =
 (0,3,-4)$.

I write out the span of $u,v,w$ and set it equal to $a,b,c$. Then I reduce that linear system to 
The last row tells me that one of my vectors was dependent on another. So I must have $\frac{1}{2}c = -\frac{2}{3}(b - \frac{1}{2}a)$. Then I solve for $c_2$ to get $c_2 = 2c_3 - \frac{2}{3}(b - \frac{1}{2}a)$. Now $c_1 = \frac{1}{2}a - c_3 + \frac{1}{3}(b - \frac{1}{2}a)$.
So, I think this is right, but how can I check? 

You row reduce the matrix
\begin{align}
\begin{bmatrix}
2 & 1 & 0 & a \\
1 & -1 & 3 & b \\
0 & 2 & -4 & c
\end{bmatrix}
&\to
\begin{bmatrix}
1 & 1/2 & 0 & a/2 \\
0 & -3/2 & 3 & b-a/2 \\
0 & 2 & -4 & c
\end{bmatrix}
\\&\to
\begin{bmatrix}
1 & 1/2 & 0 & a/2 \\
0 & 1 & -2 & -\frac{2}{3}(b-a/2) \\
0 & 2 & -4 & c
\end{bmatrix}
\\&\to
\begin{bmatrix}
1 & 1/2 & 0 & a/2 \\
0 & 1 & -2 & -\frac{2}{3}(b-a/2) \\
0 & 0 & 0 & c+\frac{4}{3}(b-a/2)
\end{bmatrix}
\end{align}
Thus the condition is
$$
c+\frac{4}{3}\left(b-\frac{a}{2}\right)=0
$$
that can be rewritten
$$
c=\frac{2}{3}a-\frac{4}{3}b
$$
No more steps are required.
Once you have a vector $(a,b,c)$ satisfying the condition, it's easy to find how you can express it in terms of the given vectors (actually of the first two, which form a basis of the span of $u,v,w$). Just compute the RREF:
$$
\begin{bmatrix}
1 & 1/2 & 0 & a/2 \\
0 & 1 & -2 & -\frac{2}{3}(b-a/2) \\
0 & 0 & 0 & 0
\end{bmatrix}
\to
\begin{bmatrix}
1 & 0 & 1 & a/2+\frac{1}{3}(b-a/2) \\
0 & 1 & -2 & -\frac{2}{3}(b-a/2) \\
0 & 0 & 0 & 0
\end{bmatrix}
$$
Thus you see that a vector satisfying the condition above can be written as
$$
\left(\frac{a}{3}+\frac{b}{3}\right)u+
\left(\frac{a}{3}-\frac{2b}{3}\right)v=\frac{a+b}{3}u+\frac{a-2b}{3}v
$$
You also see that $w=u-2v$
