Answer
$x=a$
$y=a-1$
$z=a$
Work Step by Step
Build the augmented matrix of the system of equations;
$\begin{bmatrix}1&-2&1&|&2\\2&-1&-1&|&1\end{bmatrix}$
Use Gauss-Jordan elimination:
Add $-2R_1$ to $R_2$:
$\begin{bmatrix}1&-2&1&|&2\\0&3&-3&|&-3\end{bmatrix}$
Multiply $R_2$ by $\dfrac{1}{3}$:
$\begin{bmatrix}1&-2&1&|&2\\0&1&-1&|&-1\end{bmatrix}$
Add $2R_2$ to $R_1$:
$\begin{bmatrix}1&0&-1&|&0\\0&1&-1&|&-1\end{bmatrix}$
As $z$ is a free variable, note $z=a$:
$\begin{cases}
x-z=0\\
y-z=-1
\end{cases}$
$\begin{cases}
x-a=0\\
y-a=-1
\end{cases}$
The solution is:
$x=a$
$y=a-1$
$z=a$
