The greatest value of for which the system of linear equations
`x-cy-cz=0`
`cx-y+cz=0`
`cx+cy-z=0`
Has a non-trivial solution, is `1/k`. The value of k is _________.

Explanation

To find the greatest value of \( C \) for which the system of linear equations has a non-trivial solution, we need to analyze the given equations:

1. \( x - cy - cz = 0 \)
2. \( cx - y + cz = 0 \)
3. \( cx + cy - z = 0 \)

### Step 1: Write the system in matrix form
We can represent the system of equations in matrix form as \( A \mathbf{X} = 0 \), where \( A \) is the coefficient matrix and \( \mathbf{X} \) is the vector of variables.

\[ A = \begin{bmatrix} 1 & -c & -c \\ c & -1 & c \\ c & c & -1 \end{bmatrix}, \quad \mathbf{X} = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \]

### Step 2: Find the determinant of the coefficient matrix
For the system to have a non-trivial solution, the determinant of the coefficient matrix must be zero:

\[ \text{det}(A) = 0 \]

Calculating the determinant:

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

### Step 3: Calculate the determinant using cofactor expansion
Using cofactor expansion along the first row:

\[ \text{det}(A) = 1 \cdot \begin{vmatrix} -1 & c \\ c & -1 \end{vmatrix} - (-c) \cdot \begin{vmatrix} c & c \\ c & -1 \end{vmatrix} - c \cdot \begin{vmatrix} c & -1 \\ c & c \end{vmatrix} \]

Calculating the 2x2 determinants:

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

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

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

Putting it all together:

\[ \text{det}(A) = 1 - c^2 + c(-c(1 + c)) - c(c(c + 1)) \]

This simplifies to:

\[ \text{det}(A) = 1 - c^2 - c^2 - c^3 - c^2 = 1 - 3c^2 - c^3 \]

### Step 4: Set the determinant to zero
Now we set the determinant equal to zero:

\[ 1 - 3c^2 - c^3 = 0 \]

Rearranging gives:

\[ c^3 + 3c^2 - 1 = 0 \]

### Step 5: Find the roots of the cubic equation
To find the roots, we can use the Rational Root Theorem or numerical methods. Testing \( c = 1 \):

\[ 1^3 + 3(1^2) - 1 = 1 + 3 - 1 = 3 \quad \text{(not a root)} \]

Testing \( c = -1 \):

\[ (-1)^3 + 3(-1)^2 - 1 = -1 + 3 - 1 = 1 \quad \text{(not a root)} \]

Testing \( c = \frac{1}{2} \):

\[ \left(\frac{1}{2}\right)^3 + 3\left(\frac{1}{2}\right)^2 - 1 = \frac{1}{8} + \frac{3}{4} - 1 = \frac{1}{8} + \frac{6}{8} - \frac{8}{8} = 0 \quad \text{(is a root)} \]

### Step 6: Factor the cubic equation
Since \( c = \frac{1}{2} \) is a root, we can factor \( c^3 + 3c^2 - 1 \) as:

\[ (c - \frac{1}{2})(c^2 + \frac{7}{2}c + 2) = 0 \]

### Step 7: Find the maximum value of \( C \)
The quadratic \( c^2 + \frac{7}{2}c + 2 \) can be analyzed using the discriminant:

\[ D = \left(\frac{7}{2}\right)^2 - 4 \cdot 1 \cdot 2 = \frac{49}{4} - 8 = \frac{49}{4} - \frac{32}{4} = \frac{17}{4} > 0 \]

Thus, there are two real roots from the quadratic equation, but we are interested in the maximum value of \( C \).

### Step 8: Evaluate \( \frac{1}{C} \)
The maximum value of \( C \) is \( \frac{1}{2} \), thus:

\[ \frac{1}{C} = 2 \]

### Conclusion
The value of \( k \) is:

\[ \boxed{2} \]