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(A) We know that for any square matrix $A$,$A \cdot (	ext{Adj } A) = |A| I$,where $I$ is the identity matrix of the same order.First,we calculate the

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$\begin{bmatrix} 68 & 0 & 0 \ 0 & 68 & 0 \ 0 & 0 & 68 \end{bmatrix}$

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(A) We know that for any square matrix $A$,$A \cdot (	ext{Adj } A) = |A| I$,where $I$ is the identity matrix of the same order.First,we calculate the determinant of matrix $A$:$|A| = \begin{vmatrix} x & 3 & 2 \ 1 & y & 4 \ 2 & 2 & z \end{vmatrix}$Expanding along the first row:$|A| = x(yz - 8) - 3(z - 8) + 2(2 - 2y)$$|A| = xyz - 8x - 3z + 24 + 4 - 4y$$|A| = xyz - (8x + 4y + 3z) + 28$Given $xyz = 60$ and $8x + 4y + 3z = 20$,we substitute these values:$|A| = 60 - 20 + 28 = 68$Therefore,$A \cdot (	ext{Adj } A) = 68 I = 68 \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 68 & 0 & 0 \ 0 & 68 & 0 \ 0 & 0 & 68 \end{bmatrix}$.

If matrix $A = \begin{bmatrix} x & 3 & 2 \\ 1 & y & 4 \\ 2 & 2 & z \end{bmatrix}$,$xyz = 60$ and $8x + 4y + 3z = 20$,then $A \cdot (\text{Adj } A)$ is equal to

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