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 solve the system of equations for the value of \( k \) that allows for a nontrivial solution, we need to set up the determinant of the coefficients and solve for when it equals zero. Here’s a step-by-step solution:

### Step 1: Write the System of Equations
The given 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 2: Set Up the Determinant
The determinant \( D \) of the coefficients of the system can be represented as:
\[ D = \begin{vmatrix} k & k+1 & k-1 \\ k+1 & k & k+2 \\ k-1 & k+2 & k \end{vmatrix} \]

### Step 3: Calculate the Determinant
To calculate the determinant \( D \), we can use the formula for the determinant of a 3x3 matrix:
\[ D = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where the matrix elements are:
\[ \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} \]

For our matrix:
- \( a = k \), \( b = k+1 \), \( c = k-1 \)
- \( d = k+1 \), \( e = k \), \( f = k+2 \)
- \( g = k-1 \), \( h = k+2 \), \( i = k \)

Calculating the determinant:
\[ D = k \left( k(k) - (k+2)(k+2) \right) - (k+1) \left( (k+1)(k) - (k+2)(k-1) \right) + (k-1) \left( (k+1)(k+2) - k(k-1) \right) \]

### Step 4: Expand and Simplify
Now we need to expand and simplify each term:
1. \( k(k^2 - (k^2 + 4k + 4)) = k(-4k - 4) = -4k^2 - 4k \)
2. \( -(k+1)((k^2 + k) - (k^2 + k - 2)) = -(k+1)(2) = -2k - 2 \)
3. \( (k-1)((k^2 + 3k + 2) - (k^2 - k)) = (k-1)(4k + 2) = 4k^2 + 2k - 4k - 2 = 4k^2 - 2k - 2 \)

Combining all these:
\[ D = -4k^2 - 4k - 2k - 2 + 4k^2 - 2k - 2 \]
\[ D = -8k - 4 \]

### Step 5: Set the Determinant to Zero
For a nontrivial solution, we set the determinant equal to zero:
\[ -8k - 4 = 0 \]
\[ -8k = 4 \quad \Rightarrow \quad k = -\frac{1}{2} \]

### Conclusion
The value of \( k \) for which the system has a nontrivial solution is:
\[ \boxed{-\frac{1}{2}} \]