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_(lambdainS)(|lambda|^(2)+| lambda|)` is equal to

Options

A. `12`
B. `4`
C. `2`
D. `6`

Answer

To solve the problem, we need to determine the values of \(\lambda\) for which the given system of equations is inconsistent. The system of equations is:

1. \(\lambda x + y + z = 1\)
2. \(x + \lambda y + z = 1\)
3. \(x + y + \lambda z = 1\)

### Step 1: Write the system in matrix form
We can express the system in matrix form as follows:

\[ \begin{bmatrix} \lambda & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \]

### Step 2: Find the determinant of the coefficient matrix
The system will be inconsistent if the determinant of the coefficient matrix is zero, but the augmented matrix has a different rank. We need to calculate the determinant:

\[ D = \begin{vmatrix} \lambda & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda \end{vmatrix} \]

### Step 3: Calculate the determinant
Using the formula for the determinant of a \(3 \times 3\) matrix:

\[ D = \lambda \begin{vmatrix} \lambda & 1 \\ 1 & \lambda \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & \lambda \end{vmatrix} + 1 \begin{vmatrix} 1 & \lambda \\ 1 & 1 \end{vmatrix} \]

Calculating the \(2 \times 2\) determinants:

1. \(\begin{vmatrix} \lambda & 1 \\ 1 & \lambda \end{vmatrix} = \lambda^2 - 1\)

2. \(\begin{vmatrix} 1 & 1 \\ 1 & \lambda \end{vmatrix} = \lambda - 1\)

3. \(\begin{vmatrix} 1 & \lambda \\ 1 & 1 \end{vmatrix} = 1 - \lambda\)

Substituting these back into the determinant:

\[ D = \lambda(\lambda^2 - 1) - (\lambda - 1) + (1 - \lambda) \]

Simplifying this expression:

\[ D = \lambda^3 - \lambda - \lambda + 1 + 1 - \lambda = \lambda^3 - 3\lambda + 2 \]

### Step 4: Set the determinant to zero
To find the values of \(\lambda\) for which the system is inconsistent, we set the determinant to zero:

\[ \lambda^3 - 3\lambda + 2 = 0 \]

### Step 5: Factor the polynomial
We can factor this polynomial. By testing possible rational roots, we find that \(\lambda = 1\) is a root. We can factor the polynomial as:

\[ (\lambda - 1)(\lambda^2 + \lambda - 2) = 0 \]

Factoring the quadratic:

\[ \lambda^2 + \lambda - 2 = (\lambda - 1)(\lambda + 2) \]

Thus, the complete factorization is:

\[ (\lambda - 1)^2(\lambda + 2) = 0 \]

### Step 6: Find the roots
The roots are:

1. \(\lambda = 1\) (double root)
2. \(\lambda = -2\)

### Step 7: Determine inconsistency
The system is inconsistent when \(\lambda = -2\) (as \(\lambda = 1\) leads to infinitely many solutions).

### Step 8: Calculate the required sum
Now we need to find:

\[ \sum_{\lambda \in S} (|\lambda|^2 + |\lambda|) \]

For \(\lambda = -2\):

\[ |\lambda|^2 + |\lambda| = (-2)^2 + 2 = 4 + 2 = 6 \]

### Final Answer
Thus, the required sum is:

\[ \boxed{6} \]