How to solve system of three equations I got these equations as a system 
$2x+3y-2z=2$
$3x-y+2z=-1$
$7x+16y-12z=11$
And I cannot figure out how to solve this. I have tried addition, tried to get rid of z and only solve for x and y but always end up with $0=0$

$$\left(\begin{array}{ccc|c}
2 & 3 & -2 & 2 \\
3 & -1 & 2 & -1 \\
7 & 16 & -12 & 11\\
\end{array}\right)$$
Add $-\frac{3}{2}$ times the first row to the second, and $-\frac{7}{2}$ times the first row to the third:
$$\left(\begin{array}{ccc|c}
2 & 3 & -2 & 2 \\
0 & \frac{-11}{2} & 5 & -4 \\
0 & \frac{11}{2} & -5 & 4\\
\end{array}\right)$$
Now add the second row to the third:
$$\left(\begin{array}{ccc|c}
2 & 3 & -2 & 2 \\
0 & \frac{-11}{2} & 5 & -4 \\
0 & 0 & 0 & 0\\
\end{array}\right)$$
Remove the third row:
$$\left(\begin{array}{ccc|c}
2 & 3 & -2 & 2 \\
0 & \frac{-11}{2} & 5 & -4 \\
\end{array}\right)$$
Move the third column to the right side (all of the numbers there will be multipled by $-1$), and name it as $p$ (so $z=p$)
$$\left(\begin{array}{cc|cc}
2 & 3 & 2 & 2 \\
0 & \frac{-11}{2} & -4 & -5 \\
\end{array}\right)$$
Now add $\frac{6}{11}$ times the second row to the first:
$$\left(\begin{array}{cc|cc}
2 & 0 & \frac{-2}{11} & \frac{-8}{11} \\
0 & \frac{-11}{2} & -4 & -5 \\
\end{array}\right)$$
Now the equations:
$$2x=-\frac{2}{11}-\frac{8}{11}p$$
$$-\frac{11}{2}y=-4-5p$$
$$z=p$$
So:
$$x=-\frac{1}{11}-\frac{4}{11}p$$
$$y=-\frac{8}{11}-\frac{10}{11}p$$
$$z=p$$
In vector form:
$$\begin{pmatrix} x \\ y \\ z\end{pmatrix}=\frac{1}{11}\begin{pmatrix} -1 \\ -8 \\ 0\end{pmatrix} + \frac{p}{11} \begin{pmatrix} -4 \\ -10 \\ 11\end{pmatrix}$$
