#### Answer

The solution of the system of equations is $\left( 1,-1,1 \right)$

#### Work Step by Step

It is given that:
$2x+y-2z=-1$ ...... (I)
$3x-3y-z=5$ ...... (II)
$x-2y+3z=6$ ...... (III)
Multiply equation (III) by 2 and subtract it from equation (I):
$\begin{align}
& 2x+y-2z=-1 \\
& 2x-4y+6z=1 \\
\end{align}$
Solve and get:
$5y-8z=-13$ ...... (IV)
Multiply equation (III) by $3$ and subtract from equation (II):
$\begin{align}
& 3x-3y-z=5 \\
& 3x-6y+9z=18 \\
\end{align}$
Solve and get:
$3y-10z=-13$ ...... (V)
Multiply equation (V) by $5$ and equation (IV) by $3$ and subtract each other:
$\begin{align}
& 15y-24z=-39 \\
& 15y-50z=-65 \\
\end{align}$
Solve and we get:
$\begin{align}
& 26z=26 \\
& z=1
\end{align}$
Putting the value of $z=1$ in equation (IV):
$\begin{align}
& 5y-8\left( 1 \right)=-13 \\
& 5y=-5 \\
& y=-1 \\
\end{align}$
Putting the value of $z=1$ and $y=-1$ in equation (I):
$\begin{align}
& 2x+\left( -1 \right)-2\left( 1 \right)=-1 \\
& 2x-1-2=-1 \\
& 2x=2 \\
& x=1
\end{align}$
Thus, the order triple $\left( 1,-1,1 \right)$ satisfies the three systems.