Solve the System. Give answer as $(x, y, z)$.
\[
\begin{array}{l}
\left\{\begin{array}{l}
x-y+6 z=16 \\
2 x+y-3 z=-4 \\
-4 x+y+2 z=5
\end{array}\right. \\
(x, y, z)=
\end{array}
\]

Step 1 ：Given the system of equations: \[ \begin{cases} x-y+6z=16 \ 2x+y-3z=-4 \ -4x+y+2z=5 \end{cases} \]

Step 2 ：We can write this system in matrix form as: \[ A = \begin{bmatrix} 1 & -1 & 6 \ 2 & 1 & -3 \ -4 & 1 & 2 \end{bmatrix}, B = \begin{bmatrix} 16 \ -4 \ 5 \end{bmatrix} \]

Step 3 ：Find the inverse of matrix A, denoted as \(A_{inv}\): \[ A_{inv} = \begin{bmatrix} 0.15151515 & 0.24242424 & -0.09090909 \ 0.24242424 & 0.78787879 & 0.45454545 \ 0.18181818 & 0.09090909 & 0.09090909 \end{bmatrix} \]

Step 4 ：Then, multiply the inverse of A with B to find the values of the variables x, y, and z: \[ X = A_{inv} \cdot B = \begin{bmatrix} 1 \ 3 \ 3 \end{bmatrix} \]

Step 5 ：Final Answer: The solution to the system of equations is \((x, y, z) = \boxed{(1, 3, 3)}\)