Answer
$(x,y,z)=(2,-1,1)$
Work Step by Step
To solve the given system,
\begin{align*}\require{cancel}
x+y-z&=0
&(\text{Eqn 1})
\\
3x+y+z&=6
&(\text{Eqn 2})
\\
x-y+2z&=5
&(\text{Eqn 3})
,\end{align*}
Subtract Eqn $(1)$ and Eqn $(2)$. This results to
\begin{align*}
-2x-2z&=-6
\\\\
\dfrac{-2x-2z}{-2}&=\dfrac{-6}{-2}
\\\\
x+z&=3
&(\text{Eqn 4})
.\end{align*}
Adding Eqn ($1$) and Eqn ($3$) results to
\begin{align*}
2x+z&=5
&(\text{Eqn 5})
.\end{align*}
Subtracting Eqn $(4)$ and Eqn $(5)$ results to
\begin{align*}
-x&=-2
\\
x&=2
.\end{align*}
Substituting $x=2$ in $x+z=3$ (Eqn $4$) results to
\begin{align*}
2+z&=3
\\
2-2+z&=3-2
\\
z&=1
.\end{align*}
Substituting $x=2$ and $z=1$ in $x+y-z=0$ (Eqn $1$) results to
\begin{align*}\require{cancel}
2+y-1&=0
\\
1+y&=0
\\
1-1+y&=0-1
\\
y&=-1
.\end{align*}
Hence, the solution to the given system is $(x,y,z)=(2,-1,1)$.
