Using Cramer’s rule, we get, 
$\mathrm{\Delta }=\left|\begin{array}{lll}2& -1& -1\\ -1& -1& 2\\ 1& -2& 1\end{array}\right|$
Using $C_1=C_1+C_2+C_3$
$\mathrm{\Delta }=\left|\begin{array}{lll}0& -1& -1\\ 0& -1& 2\\ 0& -2& 1\end{array}\right|=0$
${\mathrm{\Delta }}_1=\left|\begin{array}{lll}1& -1& -1\\ 1& -1& 2\\ 2& -2& 1\end{array}\right|=0$ (as $C_1$ and $C_2$ are identical)
${\mathrm{\Delta }}_2=\left|\begin{array}{lll}2& 1& -1\\ -1& 1& 2\\ 1& 2& 1\end{array}\right|$
Using $C_1=C_1-C_2+C_3$
$=\left|\begin{array}{lll}0& 1& -1\\ 0& 1& 2\\ 0& 2& 1\end{array}\right|=0$
${\mathrm{\Delta }}_3=\left|\begin{array}{lll}2& -1& 1\\ -1& -1& 1\\ 1& -2& 2\end{array}\right|=0$ (as $C_2$ and $C_3$ are identical)
Therefore, infinite solutions
Now, put $z=\lambda$, we get, 
$2x-y=1+\lambda$ and
$-x-y=1-2\lambda$
Solving these $2$ equations, we get, 
$x=\lambda , y=1-\lambda$
$\frac{x_0^2-y_0^2+1}{z_0}=\frac{{\lambda }^2-{\left(\lambda -1\right)}^2+1}{\lambda }=2$