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

We need to find the set $$S$$ of all real values of $$\lambda$$ for which the system is inconsistent:

$$\lambda x + y + z = 1$$

$$x + \lambda y + z = 1$$

$$x + y + \lambda z = 1$$

$$D = \begin{vmatrix} \lambda & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda \end{vmatrix}$$

Expanding: $$D = \lambda(\lambda^2 - 1) - 1(\lambda - 1) + 1(1 - \lambda)$$

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

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

$$= (\lambda - 1)^2(\lambda + 2)$$

$$D = 0$$ when $$\lambda = 1$$ or $$\lambda = -2$$.

The system becomes $$x + y + z = 1$$ (all three equations identical). This has infinitely many solutions (consistent). So $$\lambda = 1 \notin S$$.

The system becomes:

$$-2x + y + z = 1$$

$$x - 2y + z = 1$$

$$x + y - 2z = 1$$

Adding all three: $$0 = 3$$, which is a contradiction. So the system is inconsistent. Thus $$\lambda = -2 \in S$$.

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

$$\sum_{\lambda \in S}(\lambda^2 + \lambda) = (-2)^2 + (-2) = 4 - 2 = 2$$

The answer is Option A: $$2$$.