JEE Math Practice Question

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:

1. A.16
2. B.36
3. C.49
4. D.1