$$ B=\left[\begin{array}{lll} 1 & 3 & \alpha \\ 1 & 2 & 3 \\ \alpha & \alpha & 4 \end{array}\right], \alpha>2 $$

And $\operatorname{adj}(A)=B,|A|=2$

$$ \begin{aligned} & \Rightarrow|\operatorname{adj}(A)|=|B| \\\\ & \Rightarrow 2^2=(8-3 \alpha)-3(4-3 \alpha)+\alpha(-\alpha) \\\\ & \Rightarrow \alpha^2-6 \alpha+8=0 \end{aligned} $$

$$ \begin{aligned} \Rightarrow & (\alpha-4)(\alpha-2)=0 \\\\ & \alpha=4,2 \text { but } \alpha>2 \text { so } \alpha=4 \end{aligned} $$

Now

$$ \begin{aligned} & {\left[\begin{array}{ccc}\alpha & -2 \alpha & \alpha\end{array}\right] B\left[\begin{array}{c} \alpha \\ -2 \alpha \\ \alpha \end{array}\right]=\left[\begin{array}{lll} 4-8 & 4 \end{array}\right]\left[\begin{array}{lll} 1 & 3 & 4 \\ 1 & 2 & 3 \\ 4 & 4 & 4 \end{array}\right]\left[\begin{array}{c} 4 \\ -8 \\ 4 \end{array}\right]} \\\\ & =\left[\begin{array}{lll} 12 & 12 & 8 \end{array}\right]\left[\begin{array}{c} 4 \\ -8 \\ 4 \end{array}\right] \\\\ & = {48-96+32=-16} \end{aligned} $$