Mathematics - Determinant Question with Solution | TestHub

 For no solution $\Delta =0$ and at least one of ${\Delta }_1,{\Delta }_2,{\Delta }_3$ is non-zero.
$\Delta =\left|\begin{array}{ccc}2& -1& 2\\ 1& -2& \lambda \\ 1& \lambda & 1\end{array}\right|=-\left(\lambda -1\right)\left(2\lambda +1\right)$
${\Delta }_1=\left|\begin{array}{ccc}2& -1& 2\\ -4& -2& \lambda \\ 4& \lambda & 1\end{array}\right|=-2\left({\lambda }^2+6\lambda -4\right)$
$\Delta =0\Rightarrow \lambda =1,-\frac{1}{2}$

 And for  $\lambda =1,-\frac{1}{2}, {∆}_1\ne 0$
Hence, $S=\left\{1,-\frac{1}{2}\right\}$