Let S denote the set of all real values of λ such that the system of equations λ x + y + z = 1 x +…

System of equations 

$$\begin{gathered} \lambda x+y+z=1 \\ x+\lambda y+z=1 \\ x+y+\lambda z=1 \end{gathered}$$

has no solution, so

$\left|\begin{array}{ccc}\lambda & 1& 1\\ 1& \lambda & 1\\ 1& 1& \lambda \end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc}\lambda +2& 2+\lambda & 2+\lambda \\ 1& \lambda & 1\\ 1& 1& \lambda \end{array}\right|=0$

$\Rightarrow \left(2+\lambda \right)\left|\begin{array}{ccc}1& 1& 1\\ 1& \lambda & 1\\ 1& 1& \lambda \end{array}\right|=0$

$\Rightarrow \left(2+\lambda \right)\left|\begin{array}{ccc}1& 0& 0\\ 1& \lambda -1& 0\\ 1& 0& \lambda -1\end{array}\right|=0$

$\Rightarrow \left(2+\lambda \right){\left(\lambda -1\right)}^2=0$

$\Rightarrow \lambda =1, -2$

For $\lambda =1$, we get infinite solutions, since equation will be identical.

So,

$\sum _{}^{}\left({\left|\lambda \right|}^2+\left|\lambda \right|\right)=4+2=6$