#### Answer

$\left( 1,-1,0 \right)$ satisfies the given system of equations.

#### Work Step by Step

Substitute $z=0$ in $\left( 12z+1,10z-1,z \right)$
Thus, $\left( 1,-1,0 \right)$
Now by substituting $x=1,y=-1,z=0$ in the left-hand side of the equation (1), we get,
Thus,
$\begin{align}
& 3x-4y+4z=3\times 1-4\times -1+4\times 0 \\
& =3+4 \\
& =7
\end{align}$
The value obtained is equal to the right-hand side value of the equation.
Therefore, $\left( 1,-1,0 \right)$ satisfies equation (1).
Now by substituting $x=1,y=-1,z=0$ in the left-hand side of the equation (2), we get,
Thus,
$\begin{align}
& x-y-2z=1-1\times -1-2\times 0 \\
& =1+1 \\
& =2
\end{align}$
The value obtained is equal to the right-hand side value of the equation.
Therefore, $\left( 1,-1,0 \right)$ satisfies equation (2).
Now, by substituting $x=1,y=-1,z=0$ in the left-hand side of the equation (3), we get,
Thus,
$\begin{align}
& 2x-3y+6z=5=2\times 1-3\times -1+6\times 0 \\
& =2+3 \\
& =5
\end{align}$
The value obtained is equal to the right hand side value of the equation.
Therefore, $\left( 1,-1,0 \right)$ satisfies equation (3).
Thus, $\left( 1,-1,0 \right)$ satisfies the given system of equations.
Hence, $\left( 12z+1,10z-1,z \right)$ for $z=0$ satisfied the given system.