First row of a matrix A is  $\left[\begin{array}{ccc}1& 3& 2\end{array}\right]$. If  $adjA=\left[\begin{array}{ccc}-2& 4& \alpha \\ -1& 2& 1\\ 3\alpha & -5& -2\end{array}\right]$ then $\alpha =$

$$\begin{gathered} \left|A\right|=a_{11}{A}_{11}+a_{12}{A}_{12}+a_{13}{A}_{13} \\ =\left(1\right)(-2)+3(-1)+2\left(3\alpha \right) \\ =6\alpha -5 \end{gathered}$$

$$\begin{gathered} \left|AdjA\right|={\left|A\right|}^2 \\ \therefore {\left|A\right|}^2=-6{\alpha }^2+17\alpha -10 \\ \therefore { (6\alpha -5)}^2=-6{\alpha }^2+17\alpha -10 \\ \begin{array}{r}\Rightarrow 42{\alpha }^2-77\alpha +35=0\\ \Rightarrow 6{\alpha }^2-11\alpha +5=0\\ \Rightarrow (\alpha -1)(6\alpha -5)=0\end{array} \\ \Rightarrow \alpha =1 OR \Rightarrow \alpha =\frac{5}{6} \end{gathered}$$

But $\alpha =\frac{5}{6}$ does not satisfy given condition