Let A= [1, -2, α; α, 2, -1] and B= [2, α; -1, 2; 4, -5], α ∈ C. Then the absolute value of the sum of all values of α for which det ( AB )=0 is:

Let $\displaystyle A=\left[\begin{array}{rrr}1 & -2 & \alpha \\ \alpha & 2 & -1\end{array}\right]$ and $\displaystyle B=\left[\begin{array}{rr}2 & \alpha \\ -1 & 2 \\ 4 & -5\end{array}\right], \alpha \in \mathbf{C}$. Then the absolute value of the sum of all values of $\displaystyle \alpha$ for which $\displaystyle \operatorname{det}(\mathrm{AB})=0$ is :