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

For matrix $$A$$ (3×3): $$|A| = 2$$.

First find $$\alpha$$: $$|A| = 1(6-1) - 2(2\alpha-1) + 3(\alpha-3) = 5 - 4\alpha + 2 + 3\alpha - 9 = -\alpha - 2 = 2$$

$$\alpha = -4$$

Now: $$|2 \text{adj}(2\text{adj}(2A))| = |2|^3 \cdot |\text{adj}(2\text{adj}(2A))|$$

For an $$n \times n$$ matrix M: $$|\text{adj}(M)| = |M|^{n-1}$$ and $$|kM| = k^n|M|$$.

$$|2A| = 2^3|A| = 8 \times 2 = 16$$

$$|\text{adj}(2A)| = |2A|^2 = 256$$

$$|2\text{adj}(2A)| = 2^3 \times 256 = 2048$$

$$|\text{adj}(2\text{adj}(2A))| = |2\text{adj}(2A)|^2 = 2048^2$$

$$|2\text{adj}(2\text{adj}(2A))| = 2^3 \times 2048^2 = 8 \times 2048^2$$

$$2048 = 2^{11}$$, so $$2048^2 = 2^{22}$$

$$= 2^3 \times 2^{22} = 2^{25} = 32^5$$

So $$32^n = 32^5$$, thus $$n = 5$$.

$$3n + \alpha = 15 + (-4) = 11$$

This matches option 2: 11.