Find the values of $\alpha $ satisfying the equation $$\begin{vmatrix} (1+\alpha)^2 & (1+2\alpha)^2 & (1+3\alpha)^2\\ (2+\alpha)^2& (2+2\alpha)^2 & (2+3\alpha)^2\\ (3+\alpha)^2& (3+2\alpha)^2 & (3+3\alpha)^2 \end{vmatrix}=-648\alpha $$
I used $$R_3 \rightarrow R_3- R_2 \qquad R_2 \rightarrow R_2- R_1$$ $$\begin{vmatrix} (1+\alpha)^2 & (1+2\alpha)^2 & (1+3\alpha)^2\\ 3+2\alpha& 3+4\alpha & 3+6\alpha\\ 5+2\alpha& 5+4\alpha & 5+6\alpha \end{vmatrix}$$ Then $$R_3 \rightarrow R_3- R_2$$ $$\begin{vmatrix} (1+\alpha)^2 & (1+2\alpha)^2 & (1+3\alpha)^2\\ 3+2\alpha& 3+4\alpha & 3+6\alpha\\ 2& 2 & 2 \end{vmatrix}$$ Now applying column will make zero but the question will become too long . This is contest question and so it should not be that long .
Hint write it as a product of two determinants after taking $\alpha,\alpha^2$ common from one of the determinants to get $\alpha=\pm 9$ or to continue your method use $R_1\to R_1-R_2$