Let $\alpha \in(0, \infty)$ and $$A=\left[\begin{array}{lll}1 & 2 & \alpha \\ 1 & 0 & 1 \\ 0 & 1 & 2\end{array}\right]$$. If $$\operatorname{det}\left(\operatorname{adj}\left(2 A-A^T\right) \cdot \operatorname{adj}\left(A-2 A^T\right)\right)=2^8$$, then $(\operatorname{det}(A))^2$ is equal to:

<p>$$\begin{aligned} & \left|\operatorname{adj}\left(A-2 A^T\right) \cdot \operatorname{adj}\left(2 A-A^T\right)\right|=2^8 \\ & P=A-2 A^{\top} \\ & Q=2 A^T-A \Rightarrow Q^T=2 A^T-A=-P \\ & |\operatorname{adj}(P) \operatorname{adj}(Q)| \Rightarrow|P Q|=-2^4 \\ & \Rightarrow|P|(-|P|)=-2^4 \Rightarrow|P|=4 \text { and }|Q|=-4 \\ & \left|A-2 A^T\right|=4 \\ & A-2 A^T=\left[\begin{array}{lll} 1 & 2 & \alpha \\ 1 & 0 & 1 \\ 0 & 1 & 2 \end{array}\right]-2\left[\begin{array}{lll} 1 & 1 & 0 \\ 2 & 0 & 1 \\ \alpha & 1 & 2 \end{array}\right]=\left[\begin{array}{ccc} -1 & 0 & \alpha \\ -3 & 0 & -1 \\ -2 \alpha & -1 & -2 \end{array}\right] \\ & \Rightarrow\left|A-2 A^T\right|=1+3 \alpha=4 \Rightarrow \alpha=1 \Rightarrow|A|=-4 \\ & \Rightarrow|A|^2=16 \end{aligned}$$</p>