#### Answer

The solution of the system is $\left( 1,1,2 \right)$.

#### Work Step by Step

Let us consider the given equation:
$x+z=3$ ...... (I)
$x+2y-z=1$ ...... (II)
$2x-y+z=3$ ...... (III)
Now, multiply equation (III) by 2 and add equation (II):
$\begin{align}
& x+2y-z=1 \\
& 4x-2y+2z=6
\end{align}$
$5x+z=7$ ...... (IV)
And subtract equation (I) from equation (IV):
$\begin{align}
& 5x+z=7 \\
& x+z=3 \\
\end{align}$
Solve and we get:
$\begin{align}
& 4x=4 \\
& x=1
\end{align}$
Putting the value of $x=1$ in equation (I):
$\begin{align}
& 1+z=3 \\
& z=2
\end{align}$
Now, putting the value of $x=1$ and $z=2$ in equation (II):
$\begin{align}
& 1+2y-2=1 \\
& 2y=2 \\
& y=1
\end{align}$
Thus, the ordered triple $\left( 1,1,2 \right)$ satisfies the three systems.