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(A) First,we perform the matrix multiplication on the right side of the equation:$\begin{bmatrix} 4 & -2 \ 0 & -6 \ -1 & 2 \end{bmatrix} \begin{bmat

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$(-4, 2, 2)$

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(A) First,we perform the matrix multiplication on the right side of the equation:$\begin{bmatrix} 4 & -2 \ 0 & -6 \ -1 & 2 \end{bmatrix} \begin{bmatrix} 2 \ 1 \end{bmatrix} = \begin{bmatrix} 4(2) + (-2)(1) \ 0(2) + (-6)(1) \ -1(2) + 2(1) \end{bmatrix} = \begin{bmatrix} 8 - 2 \ 0 - 6 \ -2 + 2 \end{bmatrix} = \begin{bmatrix} 6 \ -6 \ 0 \end{bmatrix}$Now,we equate this to the left side:$\begin{bmatrix} 1 & 2 & 3 \ 3 & 1 & 2 \ 2 & 3 & 1 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 6 \ -6 \ 0 \end{bmatrix}$This gives us the system of linear equations:$1) \ x + 2y + 3z = 6$$2) \ 3x + y + 2z = -6$$3) \ 2x + 3y + z = 0$Adding the three equations:$(x + 3x + 2x) + (2y + y + 3y) + (3z + 2z + z) = 6 - 6 + 0$$6x + 6y + 6z = 0 \implies x + y + z = 0$From equation $(3)$,$z = -2x - 3y$. Substituting into $(1)$:$x + 2y + 3(-2x - 3y) = 6 \implies x + 2y - 6x - 9y = 6 \implies -5x - 7y = 6$Solving the system,we find $x = -4, y = 2, z = 2$.

If $\begin{bmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 4 & -2 \\ 0 & -6 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 2 \\ 1 \end{bmatrix}$,then $(x, y, z) = $

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