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(D) The system of equations is inconsistent if the determinant of the coefficient matrix $D = 0$ and at least one of the Cramer's rule determinants $(

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$6$

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(D) The system of equations is inconsistent if the determinant of the coefficient matrix $D = 0$ and at least one of the Cramer's rule determinants $(D_x, D_y, D_z)$ is non-zero.The coefficient matrix is $A = \begin{bmatrix} \lambda & 1 & 1 \ 1 & \lambda & 1 \ 1 & 1 & \lambda \end{bmatrix}$.Calculating the determinant $D = \lambda(\lambda^2 - 1) - 1(\lambda - 1) + 1(1 - \lambda) = \lambda(\lambda - 1)(\lambda + 1) - (\lambda - 1) - (\lambda - 1) = (\lambda - 1)(\lambda^2 + \lambda - 2) = (\lambda - 1)(\lambda + 2)(\lambda - 1) = (\lambda - 1)^2(\lambda + 2)$.Setting $D = 0$,we get $\lambda = 1$ or $\lambda = -2$.If $\lambda = 1$,the system becomes $x + y + z = 1$,which represents a plane,and all three equations are identical,leading to infinitely many solutions (consistent).If $\lambda = -2$,the system becomes $-2x + y + z = 1$,$x - 2y + z = 1$,and $x + y - 2z = 1$. Adding these three equations gives $0 = 3$,which is a contradiction,so the system is inconsistent.Thus,$S = \{-2\}$.The sum $\sum_{\lambda \in S} (|\lambda|^2 + |\lambda|) = (|-2|^2 + |-2|) = 4 + 2 = 6$.

Let $S$ denote the set of all real values of $\lambda$ such that the system of equations $\lambda x + y + z = 1$,$x + \lambda y + z = 1$,and $x + y + \lambda z = 1$ is inconsistent. Then,$\sum_{\lambda \in S} (|\lambda|^2 + |\lambda|)$ is equal to

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