If $\mathrm{A}=\left[\begin{array}{ccc}\mathrm{x}& 1& 2\\ 2& 4& \mathrm{x}\\ -3& 3& 2\end{array}\right]$ is a singular matrix and distinct values of x are  ${\mathrm{x}}_1,{\mathrm{x}}_2\text{ then }{\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_1{\mathrm{x}}_2=$

$\left|\mathrm{A}\right|=0$
$\left|\begin{array}{ccc}\mathrm{x}& 1& 2\\ 2& 4& \mathrm{x}\\ -3& 3& 2\end{array}\right|=0$
$\begin{array}{l}\Rightarrow \mathrm{x}(8-3\mathrm{x})-1(4+3\mathrm{x})+2(6+12)=0\\ \Rightarrow 8\mathrm{x}-3{\mathrm{x}}^2-4-3\mathrm{x}+36=0\\ \Rightarrow 3{\mathrm{x}}^2-5\mathrm{x}-32=0\\ {\mathrm{x}}_1+{\mathrm{x}}_2=\frac{5}{3},{\mathrm{x}}_1{\mathrm{x}}_2=\frac{-32}{3}\\ {\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_1{\mathrm{x}}_2=\frac{5-32}{3}=\frac{-27}{3}=-9\end{array}$