Let $$\alpha \in (0,\infty)$$ and $$A = \begin{bmatrix}1 & 2 & \alpha\\ 1 & 0 & 1\\ 0 & 1 & 2\end{bmatrix}$$. If $$\det(\text{adj}(2A-A^T)\cdot\text{adj}(A-2A^T)) = 2^8$$, then $$(\det(A))^2$$ is equal to:

$$A^T = \begin{bmatrix}1 & 1 & 0\\ 2 & 0 & 1\\ \alpha & 1 & 2\end{bmatrix}$$

 $$2A - A^T = \begin{bmatrix}1 & 3 & 2\alpha\\ 0 & 0 & 1\\ -\alpha & 1 & 2\end{bmatrix}$$

$$\det(2A - A^T) = 1(0-1) - 3(0+\alpha) + 2\alpha(0) = -1 - 3\alpha$$

Computing $$A - 2A^T = \begin{bmatrix}-1 & 0 & \alpha\\ -3 & 0 & -1\\ -2\alpha & -1 & -2\end{bmatrix}$$

$$\det(A - 2A^T) = -1(0-1) - 0 + \alpha(3-0) = 1 + 3\alpha$$

For a $$3 \times 3$$ matrix $$M$$: $$\det(\text{adj}(M)) = (\det M)^2$$. Therefore:

$$\det(\text{adj}(2A-A^T) \cdot \text{adj}(A-2A^T)) = (-1-3\alpha)^2 \cdot (1+3\alpha)^2 = (1+3\alpha)^4$$

Setting this equal to $$2^8 = 256$$:

$$(1+3\alpha)^4 = 256 \Rightarrow (1+3\alpha)^2 = 16 \Rightarrow 1+3\alpha = 4 \Rightarrow \alpha = 1$$

Therefore $$(\det A)^2 = (1 - 5)^2 = 16$$.