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(D) Given $A = \begin{bmatrix} 1 & 2 & 3 \ a & 3 & 1 \ 1 & 1 & 2 \end{bmatrix}$ and $|A| = 2$.Expanding the determinant along the first row:$|A| = 1

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$11$

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(D) Given $A = \begin{bmatrix} 1 & 2 & 3 \ a & 3 & 1 \ 1 & 1 & 2 \end{bmatrix}$ and $|A| = 2$.Expanding the determinant along the first row:$|A| = 1(6 - 1) - 2(2a - 1) + 3(a - 3) = 2$$5 - 4a + 2 + 3a - 9 = 2$$-a - 2 = 2 \implies a = -4$.Now,we evaluate $|2 \operatorname{adj}(2 \operatorname{adj}(2A))|$.Since $A$ is a $3 	imes 3$ matrix,$|kA| = k^3|A|$ and $|\operatorname{adj}(M)| = |M|^{3-1} = |M|^2$.Let $M = 2A$,then $|M| = 2^3|A| = 8(2) = 16$.$|2 \operatorname{adj}(2 \operatorname{adj}(M))| = 2^3 |\operatorname{adj}(2 \operatorname{adj}(M))| = 8 |2 \operatorname{adj}(M)|^2 = 8 \cdot (2^3 |\operatorname{adj}(M)|)^2 = 8 \cdot 8^2 \cdot |\operatorname{adj}(M)|^2 = 8^3 \cdot (|M|^2)^2 = 8^3 \cdot |M|^4$.Substituting $|M| = 16 = 2^4$:$|2 \operatorname{adj}(2 \operatorname{adj}(2A))| = (2^3)^3 \cdot (2^4)^4 = 2^9 \cdot 2^{16} = 2^{25} = (2^5)^5 = 32^5$.Thus,$n = 5$.We need to find $3n + \alpha$. Here $\alpha = a = -4$.$3(5) + (-4) = 15 - 4 = 11$.

Let $A = \begin{bmatrix} 1 & 2 & 3 \\ a & 3 & 1 \\ 1 & 1 & 2 \end{bmatrix}$ and $|A| = 2$. If $|2 \operatorname{adj}(2 \operatorname{adj}(2 A))| = 32^n$,then $3n + \alpha$ is equal to:

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