The system of equations
`{:(kx+(k+1)y+(k-1)z=0),((k+1)x+ky+(k+2)z=0),((k-1)x + (k+2)y+kz=0):}`
has a nontrivial solution for :

Options

A. Exactly three real value of k
B. Exactly two real values of k
C. Exactly one real value of k
D. Infinite number of values of k

Answer

To find the values of \( k \) for which the given system of equations has a nontrivial solution, we need to set up the determinant of the coefficient matrix and solve for when this determinant equals zero.

The system of equations is:

1. \( kx + (k+1)y + (k-1)z = 0 \)
2. \( (k+1)x + ky + (k+2)z = 0 \)
3. \( (k-1)x + (k+2)y + kz = 0 \)

### Step 1: Write the Coefficient Matrix
The coefficient matrix \( A \) for the system can be written as:

\[ A = \begin{bmatrix} k & k+1 & k-1 \\ k+1 & k & k+2 \\ k-1 & k+2 & k \end{bmatrix} \]

### Step 2: Calculate the Determinant
We need to compute the determinant of the matrix \( A \):

\[ \text{det}(A) = \begin{vmatrix} k & k+1 & k-1 \\ k+1 & k & k+2 \\ k-1 & k+2 & k \end{vmatrix} \]

### Step 3: Apply Row Operations
We will perform row operations to simplify the determinant calculation.

1. Replace \( R_2 \) with \( R_2 - R_1 \):
\[ R_2 = (k+1 - k, k - (k+1), k+2 - (k-1)) = (1, -1, 3) \]

2. Replace \( R_3 \) with \( R_3 - R_1 \):
\[ R_3 = (k-1 - k, k+2 - (k+1), k - (k-1)) = (-1, 1, 1) \]

Now the matrix looks like:

\[ \begin{bmatrix} k & k+1 & k-1 \\ 1 & -1 & 3 \\ -1 & 1 & 1 \end{bmatrix} \]

### Step 4: Calculate the Determinant
Now we compute the determinant of the modified matrix:

\[ \text{det}(A) = k \begin{vmatrix} -1 & 3 \\ 1 & 1 \end{vmatrix} - (k+1) \begin{vmatrix} 1 & 3 \\ -1 & 1 \end{vmatrix} + (k-1) \begin{vmatrix} 1 & -1 \\ -1 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} -1 & 3 \\ 1 & 1 \end{vmatrix} = (-1)(1) - (3)(1) = -1 - 3 = -4 \)

2. \( \begin{vmatrix} 1 & 3 \\ -1 & 1 \end{vmatrix} = (1)(1) - (3)(-1) = 1 + 3 = 4 \)

3. \( \begin{vmatrix} 1 & -1 \\ -1 & 1 \end{vmatrix} = (1)(1) - (-1)(-1) = 1 - 1 = 0 \)

Putting it all together:

\[ \text{det}(A) = k(-4) - (k+1)(4) + (k-1)(0) \]
\[ = -4k - 4k - 4 = -8k - 4 \]

### Step 5: Set the Determinant to Zero
For a nontrivial solution, we set the determinant equal to zero:

\[ -8k - 4 = 0 \]

### Step 6: Solve for \( k \)
Solving for \( k \):

\[ -8k = 4 \implies k = -\frac{1}{2} \]

### Final Answer
The value of \( k \) for which the system has a nontrivial solution is:

\[ k = -\frac{1}{2} \]