$\text{ Let }P=\left[\begin{array}{ccc}3& -1& -2\\ 2& 0& \alpha \\ 3& -5& 0\end{array}\right],\text{ where }\alpha \in \mathrm{ℝ}.\text{ Suppose }Q=\left[q_{ij}\right]\text{ is a matrix such that }PQ=kI,\text{ where }$$k\in \mathrm{ℝ},k\ne 0\text{ and }\mathrm{I}\text{ is the identity matrix of order }3.\text{ If }{q}_{23}=\frac{-k}{8}\text{ and }det\left(Q\right)=\frac{k^2}{2},\text{ then }$

$PQ=KI\Rightarrow \left|P\right|\left|Q\right|\ne 0$

$\Rightarrow Q=K{P}^{-1}$

$\left|P\right|=12\alpha +20$

$\therefore \frac{K^2}{2}=K^3\frac{1}{12\alpha +20}$

$Q=K{P}^{-1}=\frac{K}{12\alpha +20}\left[\begin{array}{ccc}5\alpha & 10& -\alpha \\ 3\alpha & 6& -3\alpha -4\\ -10& 12& 2\end{array}\right]$

$Q_{23}=\frac{-k}{8}=\frac{-k(3\alpha +4)}{12\alpha +20}$

$\Rightarrow \alpha =-1,k=4$

$\left|P\right(AdjQ\left)\right|=|P\Vert Q{|}^2$     $\left(\sin ce \left|Q\right|=\left|k{P}^{-1}\right|=k^3\left|P^{-1}\right|=\frac{k^3}{\left|P\right|}\right)$

$=\frac{K^6}{\left|P\right|}=2^9$

$\left|Q\left(AdjP\right)\right|=\left|Q\right|{\left|P\right|}^2=K^3\left|P\right|=2^9$