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

The coefficient matrix of the system is:

$$A = \begin{bmatrix} \lambda & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda \end{bmatrix}$$

Computing $$\det(A) = \lambda(\lambda^2 - 1) - 1(\lambda - 1) + 1(1 - \lambda)$$

$$= \lambda^3 - \lambda - \lambda + 1 + 1 - \lambda = \lambda^3 - 3\lambda + 2$$

Factoring: $$\lambda^3 - 3\lambda + 2 = (\lambda - 1)^2(\lambda + 2)$$

So $$\det(A) = 0$$ when $$\lambda = 1$$ or $$\lambda = -2$$.

For $$\lambda = 1$$: All three equations become $$x + y + z = 1$$, which has infinitely many solutions. So the system is consistent.

For $$\lambda = -2$$: Adding all three equations gives $$0 = 3$$, which is a contradiction. So the system is inconsistent.

Therefore $$S = \{-2\}$$.

$$\sum_{\lambda \in S} |\lambda|^2 + |\lambda| = |-2|^2 + |-2| = 4 + 2 = 6$$

The answer is $$6$$, which is Option D.