Let $$A=\left[\begin{array}{ccc}2 & 1 & 2 \\ 6 & 2 & 11 \\ 3 & 3 & 2\end{array}\right]$$ and $$P=\left[\begin{array}{lll}1 & 2 & 0 \\ 5 & 0 & 2 \\ 7 & 1 & 5\end{array}\right]$$. The sum of the prime factors of $$\left|P^{-1} A P-2 I\right|$$ is equal to

$$\text { Let } A=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & \alpha & \beta \\ 0 & \beta & \alpha \end{array}\right] \text { and }|2 \mathrm{~A}|^3=2^{21} \text { where } \alpha, \beta \in Z \text {, Then a value of } \alpha \text { is }$$

Let $$\mathrm{A}$$ be a square matrix such that $$\mathrm{AA}^{\mathrm{T}}=\mathrm{I}$$. Then $$\frac{1}{2} A\left[\left(A+A^T\right)^2+\left(A-A^T\right)^2\right]$$ is equal to

The values of $$\alpha$$, for which $$\left|\begin{array}{ccc}1 & \frac{3}{2} & \alpha+\frac{3}{2} \\ 1 & \frac{1}{3} & \alpha+\frac{1}{3} \\ 2 \alpha+3 & 3 \alpha+1 & 0\end{array}\right|=0$$, lie in the interval