For What $a$ The Linear Equations Have Sloutions $2x+ay-z=-2$

$x-3z=-3$

$x+2y+az=-1$
I have thought about reducing a matrix so in the end I will have an equation with $a$ then I can determine for which $a$ the are one solution/infinite solutions/no solution, but it does not seems to work
I have row reduced the matrix and got
$$\left(\begin{array}{ccc|c} 
1 & 2 & a & -1 \\
0 & -2 & 3-a & 4 \\
0 & a-4 & -a-1& -1 
\end{array}\right)$$  
the solutions are for one solution: $a=5$,$a\neq 2$  no solution for: $a=-5,a=2$

I got the Row Reduced Form as: 
$ \left( \begin{array}{ccc|c}
1 & 2 & a & -1 \\
0 & 1 & \frac{a+3}{2} & 1 \\
0 & a-5 & \frac{-5(a+1)}{2} & -1 \end{array} \right)$
So the determinant comes as $(-3a+10-a^2)$
For no solution, determinant is zero, so, $(-3a+10-a^2)=0$
$\Rightarrow a=-5, a=2$
