The system of equations:
`x+ky+3z=0`, `3x+ky-2z=0`, `2x+3y-4z=0` has non- trivial solution. when `k=`

Answer

To find the value of \( k \) for which the system of equations has a non-trivial solution, we need to analyze the determinant of the coefficient matrix. The system of equations is given as:

1. \( x + ky + 3z = 0 \)
2. \( 3x + ky - 2z = 0 \)
3. \( 2x + 3y - 4z = 0 \)

### Step 1: Form the Coefficient Matrix
The coefficient matrix \( A \) can be formed from the coefficients of \( x \), \( y \), and \( z \) in the equations:

\[ A = \begin{bmatrix} 1 & k & 3 \\ 3 & k & -2 \\ 2 & 3 & -4 \end{bmatrix} \]

### Step 2: Calculate the Determinant of the Coefficient Matrix
To find the value of \( k \) for which there is a non-trivial solution, we need to set the determinant of matrix \( A \) equal to zero:

\[ \text{det}(A) = \begin{vmatrix} 1 & k & 3 \\ 3 & k & -2 \\ 2 & 3 & -4 \end{vmatrix} \]

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

\[ \text{det}(A) = 1 \cdot \begin{vmatrix} k & -2 \\ 3 & -4 \end{vmatrix} - k \cdot \begin{vmatrix} 3 & -2 \\ 2 & -4 \end{vmatrix} + 3 \cdot \begin{vmatrix} 3 & k \\ 2 & 3 \end{vmatrix} \]

Calculating each of these \( 2 \times 2 \) determinants:

1. \( \begin{vmatrix} k & -2 \\ 3 & -4 \end{vmatrix} = k(-4) - (-2)(3) = -4k + 6 \)

2. \( \begin{vmatrix} 3 & -2 \\ 2 & -4 \end{vmatrix} = 3(-4) - (-2)(2) = -12 + 4 = -8 \)

3. \( \begin{vmatrix} 3 & k \\ 2 & 3 \end{vmatrix} = 3(3) - k(2) = 9 - 2k \)

Putting it all together:

\[ \text{det}(A) = 1(-4k + 6) - k(-8) + 3(9 - 2k) \]
\[ = -4k + 6 + 8k + 27 - 6k \]
\[ = (-4k + 8k - 6k) + (6 + 27) \]
\[ = -2k + 33 \]

### Step 4: Set the Determinant to Zero
For a non-trivial solution, we set the determinant equal to zero:

\[ -2k + 33 = 0 \]

### Step 5: Solve for \( k \)
Rearranging the equation gives:

\[ -2k = -33 \]
\[ k = \frac{33}{2} \]

### Final Answer
Thus, the value of \( k \) for which the system of equations has a non-trivial solution is:

\[ \boxed{\frac{33}{2}} \]