`k x + 3y = k - 3, 12 x + ky = k `

Explanation

To find the value of \( k \) for which the given equations do not have a solution, we start with the two equations:

1. \( kx + 3y = k - 3 \) (Equation 1)
2. \( 12x + ky = k \) (Equation 2)

### Step 1: Identify the conditions for no solution
For a system of linear equations to have no solution, the ratios of the coefficients must be equal while the ratio of the constants must be different. This can be expressed as:

\[ \frac{k}{12} = \frac{3}{k} \quad \text{(1)} \]
\[ \frac{k - 3}{k} \quad \text{(2)} \]

### Step 2: Set up the equation from the first ratio
From the first ratio, we can cross-multiply:

\[ k^2 = 36 \quad \text{(from } k \cdot k = 12 \cdot 3\text{)} \]

### Step 3: Solve for \( k \)
Taking the square root of both sides gives us:

\[ k = 6 \quad \text{or} \quad k = -6 \]

### Step 4: Check the second ratio
Now we need to ensure that these values of \( k \) do not satisfy the second ratio. We will check both values:

1. **For \( k = 6 \)**:
\[ \frac{k - 3}{k} = \frac{6 - 3}{6} = \frac{3}{6} = \frac{1}{2} \]
\[ \frac{3}{k} = \frac{3}{6} = \frac{1}{2} \]
Here, both ratios are equal, which means there is a solution.

2. **For \( k = -6 \)**:
\[ \frac{k - 3}{k} = \frac{-6 - 3}{-6} = \frac{-9}{-6} = \frac{3}{2} \]
\[ \frac{3}{k} = \frac{3}{-6} = -\frac{1}{2} \]
Here, the ratios are not equal, indicating that there is no solution.

### Conclusion
Thus, the value of \( k \) for which the given equations do not have a solution is:

\[ \boxed{-6} \]