Let S denote the set of all real values of λ such that the system...

System of equations

$\lambdax + y + z = 1x + \lambday + z = 1x + y + \lambdaz = 1$

has no solution, so

\left|$\begin{cases}\lambda & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda\end{cases}$\right| = 0$</p><p>$⇒ \left|$\begin{cases}\lambda + 2 & 2 + \lambda & 2 + \lambda \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda\end{cases}$\right| = 0$</p><p>$⇒ \left(2 + \lambda\right)\left|$\begin{cases}1 & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda\end{cases}$\right| = 0$</p><p>$⇒ \left(2 + \lambda\right)\left|$\begin{cases}1 & 0 & 0 \\ 1 & \lambda - 1 & 0 \\ 1 & 0 & \lambda - 1\end{cases}$\right| = 0$</p><p>$⇒ \left(2 + \lambda\right){\left(\lambda - 1\right)}^{2} = 0$</p><p>$⇒ \lambda = 1, - 2$</p><p>For$\lambda = 1$, we get infinite solutions, since equation will be identical.</p><p>So,</p><p>$\mathop{∑}_{}^{}\left({\left|\lambda\right|}^{2} + \left|\lambda\right|\right) = 4 + 2 = 6