We get $\left|\begin{array}{rrr}(1+\alpha)^2 & (1+2 \alpha)^2 & (1+3 \alpha)^2 \\ 3+2 \alpha & 3+4 \alpha & 3+6 \alpha \\ 5+2 \alpha & 5+4 \alpha & 5+6 \alpha\end{array}\right|=-648 \alpha \quad\left(R_3 \rightarrow R_3-R_2 ; R_2 \rightarrow R_2-R_1\right)$

$\Rightarrow\left|\begin{array}{lll}\alpha^2-2 & 4 \alpha^2-2 & 9 \alpha^2-2 \\ 3+2 \alpha & 3+4 \alpha & 3+6 \alpha \\ 2 & 2 & 2\end{array}\right|=-648 \alpha\left( R _1 \rightarrow R _1- R _2 ; R _3 \rightarrow R _3- R _2\right)$

$\Rightarrow\left|\begin{array}{ccc}-3 \alpha^2 & -5 \alpha^2 & 9 \alpha^2-3 \\ -2 \alpha & -2 \alpha & 3+6 \alpha \\ 0 & 0 & 2\end{array}\right|=-648 \alpha$

$\Rightarrow-8 \alpha^3=-648 \alpha \Rightarrow \alpha= \pm 9$

Alternate solution

$\Delta=\left|\begin{array}{lll}1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 3 & 9\end{array}\right|\left|\begin{array}{ccc}1 & 2 \alpha & \alpha^2 \\ 4 & 4 \alpha & \alpha^2 \\ 9 & 6 \alpha & \alpha^2\end{array}\right|=2 \alpha^3\left|\begin{array}{lll}1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 3 & 9\end{array}\right|\left|\begin{array}{lll}1 & 1 & 1 \\ 4 & 2 & 1 \\ 9 & 3 & 1\end{array}\right|=-2 \alpha^3\left|\begin{array}{lll}1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 3 & 9\end{array}\right|\left|\begin{array}{lll}1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 3 & 9\end{array}\right|=-2 \alpha^3 \times 4$

$\Rightarrow-8 \alpha^3=-648 \alpha \Rightarrow \alpha= \pm 9$