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(C) We know that for any square matrix $A$,the property $A (adj A) = |A| I$ holds,where $I$ is the identity matrix of the same order.First,we calculat

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

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(C) We know that for any square matrix $A$,the property $A (adj A) = |A| I$ holds,where $I$ is the identity matrix of the same order.First,we calculate the determinant $|A|$ of the matrix $A = \begin{bmatrix} x & 3 & 2 \ 1 & y & 4 \ 2 & 2 & z \end{bmatrix}$.$|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 into the expression for $|A|$:$|A| = 60 - 20 + 28 = 68$.Therefore,$A (adj A) = |A| I = 68 I = \begin{bmatrix} 68 & 0 & 0 \ 0 & 68 & 0 \ 0 & 0 & 68 \end{bmatrix}$.

Matrix $A = \begin{bmatrix} x & 3 & 2 \\ 1 & y & 4 \\ 2 & 2 & z \end{bmatrix}$. If $xyz = 60$ and $8x + 4y + 3z = 20$,then $A (adj A)$ is equal to:

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