Step 1 — Compute the product $QP$ row by row.
Row 1 of Q $= [2,\ 2,\ -4]$:
$$\begin{aligned}
(QP)_{11} &= 2(1) + 2(2) + (-4)(0) = 2 + 4 + 0 = 6\\
(QP)_{12} &= 2(-1) + 2(3) + (-4)(1) = -2 + 6 - 4 = 0\\
(QP)_{13} &= 2(0) + 2(4) + (-4)(2) = 0 + 8 - 8 = 0
\end{aligned}$$
Row 2 of Q $= [-4,\ 2,\ -4]$:
$$\begin{aligned}
(QP)_{21} &= -4(1) + 2(2) + (-4)(0) = -4 + 4 + 0 = 0\\
(QP)_{22} &= -4(-1) + 2(3) + (-4)(1) = 4 + 6 - 4 = 6\\
(QP)_{23} &= -4(0) + 2(4) + (-4)(2) = 0 + 8 - 8 = 0
\end{aligned}$$
Row 3 of Q $= [2,\ -1,\ 5]$:
$$\begin{aligned}
(QP)_{31} &= 2(1) + (-1)(2) + 5(0) = 2 - 2 + 0 = 0\\
(QP)_{32} &= 2(-1) + (-1)(3) + 5(1) = -2 - 3 + 5 = 0\\
(QP)_{33} &= 2(0) + (-1)(4) + 5(2) = 0 - 4 + 10 = 6
\end{aligned}$$
Step 2 — Assemble the result.
$$QP = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6\end{bmatrix} = 6I$$
Step 3 — Deduce $P^{-1}$.
$$QP = 6I \quad \Longrightarrow \quad \left(\frac{1}{6}Q\right)P = I \quad \Longrightarrow \quad P^{-1} = \frac{1}{6}Q$$
Step 4 — Write the system in matrix form.
$$x - y + 0z = 3, \qquad 2x + 3y + 4z = 17, \qquad 0x + y + 2z = 7$$
The coefficient matrix is exactly $P$, so
$$PX = B, \qquad X = \begin{bmatrix} x \\ y \\ z\end{bmatrix}, \qquad B = \begin{bmatrix} 3 \\ 17 \\ 7\end{bmatrix}$$
Step 5 — Solve using $X = P^{-1}B = \dfrac{1}{6}QB$.
$$QB = \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5\end{bmatrix}\begin{bmatrix} 3 \\ 17 \\ 7\end{bmatrix}$$
$$\begin{aligned}
\text{Row 1} &= 2(3) + 2(17) - 4(7) = 6 + 34 - 28 = 12\\
\text{Row 2} &= -4(3) + 2(17) - 4(7) = -12 + 34 - 28 = -6\\
\text{Row 3} &= 2(3) - 1(17) + 5(7) = 6 - 17 + 35 = 24
\end{aligned}$$
Step 6 — Multiply by $\dfrac{1}{6}$.
$$X = \frac{1}{6}\begin{bmatrix} 12 \\ -6 \\ 24\end{bmatrix} = \begin{bmatrix} 2 \\ -1 \\ 4\end{bmatrix}$$
Step 7 — Verify the solution.
$x - y = 2 - (-1) = 3$ ✓
$2(2) + 3(-1) + 4(4) = 4 - 3 + 16 = 17$ ✓
$y + 2z = -1 + 8 = 7$ ✓
$$\boxed{x = 2,\quad y = -1,\quad z = 4}$$
