#### Answer

The solution of the system is $\left( 1,0,-3 \right)$.

#### Work Step by Step

Let us consider the given equation:
$2x+y=2$ …… (I)
$x+y-z=4$ …… (II)
$3x+2y+z=0$ …… (III)
Now, adding equation (II) and equation(III):
$\begin{align}
& x+y-z=4 \\
& 3x+2y+z=0 \\
& \overline{\,\,\,\,\,4x+3y=4} \\
\end{align}$
$4x+3y=4$ …… (IV)
Equation (I) is multiplied by 2 and we subtract from equation(IV):
$\begin{align}
& 4x+3y=4 \\
& 4x+2y=4 \\
& \overline{\,\,\,\,\,\,\,\,\,\,\,\,\,\,y=0\,\,\,\,\,} \\
\end{align}$
Now, putting the value of $y=0$ in equation (I):
$\begin{align}
& 2x+0=2 \\
& x=1\, \\
\end{align}$
$x=1,y=0$
Putting in equation (III):
$\begin{align}
& 3\left( 1 \right)+2(0)+z=0 \\
& 3+0+z=0 \\
& z=-3
\end{align}$
Thus, the ordered triple $\left( 1,0,-3 \right)$ satisfies the three equations.