#### Answer

$\left( 13,9,1 \right)$ satisfies the given system.

#### Work Step by Step

Substitute $z=1$ in $\left( 12z+1,10z-1,z \right)$
Thus, $\left( 13,9,1 \right)$
Now by substituting $x=13,y=9,z=1$ in the left-hand side of the equation (I), we get,
Thus,
$\begin{align}
& 3x-4y+4z=3\times 13-4\times 9+4\times 1 \\
& =39-36+4 \\
& =7
\end{align}$
The value obtained is equal to the right-hand side value of the equation.
Hence, $\left( 13,9,1 \right)$ satisfies equation (I).
Now, by substituting $x=13,y=9,z=1$ in the left-hand side of the equation (II), we get,
Thus,
$\begin{align}
& x-y-2z=13-9-2\times 1 \\
& =13-9-2 \\
& =2
\end{align}$
The value obtained is equal to the right hand side value of the equation.
Hence, $\left( 13,9,1 \right)$ satisfies equation (II).
Now by substituting $x=13,y=9,z=1$ in the left-hand side of the equation (III), we get,
Thus,
$\begin{align}
& 2x-3y+6z=2\times 13-3\times 9+6\times 1 \\
& =26-27+6 \\
& =5
\end{align}$
The value obtained is equal to the right hand side value of the equation.
Therefore, $\left( 13,9,1 \right)$ satisfies equation (III).
Thus, $\left( 13,9,1 \right)$ satisfies the given system.
Hence, $\left( 12z+1,10z-1,z \right)$ for $z=1$ satisfied the given system.