STATEMENT-1 : The system of equations x + ky + 3z =0, 3x + ky - 2z =0, 2x+3y-z=0, possesses a non-trival solution.
then value of k is `31/2`
STATEMENT -2 Three linear equations in x, y, z can never have no solution if it is homogeneous, hence exactly two types of possible solution.

Answer

To determine the value of \( k \) for which the system of equations has a non-trivial solution, we need to analyze the given homogeneous system of equations:

1. **Equations:**
\[ \begin{align*} x + ky + 3z &= 0 \quad \text{(1)} \\ 3x + ky - 2z &= 0 \quad \text{(2)} \\ 2x + 3y - z &= 0 \quad \text{(3)} \end{align*} \]

2. **Matrix Representation:**
We can represent this system in matrix form as:
\[ A = \begin{bmatrix} 1 & k & 3 \\ 3 & k & -2 \\ 2 & 3 & -1 \end{bmatrix} \]

3. **Condition for Non-Trivial Solution:**
For the system to have a non-trivial solution, the determinant of the coefficient matrix \( A \) must be equal to zero:
\[ \text{det}(A) = 0 \]

4. **Calculating the Determinant:**
We will compute the determinant of matrix \( A \):
\[ \text{det}(A) = \begin{vmatrix} 1 & k & 3 \\ 3 & k & -2 \\ 2 & 3 & -1 \end{vmatrix} \]

Using the formula for the determinant of a 3x3 matrix:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where \( a, b, c \) are the elements of the first row, and \( d, e, f, g, h, i \) are the elements of the remaining rows.

Substituting the values:
\[ = 1 \cdot (k \cdot (-1) - (-2) \cdot 3) - k \cdot (3 \cdot (-1) - (-2) \cdot 2) + 3 \cdot (3 \cdot 3 - k \cdot 2) \]

Simplifying each term:
\[ = 1 \cdot (-k + 6) - k \cdot (-3 + 4) + 3 \cdot (9 - 2k) \]
\[ = -k + 6 + k + 3 \cdot (9 - 2k) \]
\[ = -k + 6 + 27 - 6k \]
\[ = 33 - 7k \]

5. **Setting the Determinant to Zero:**
To find \( k \):
\[ 33 - 7k = 0 \]
\[ 7k = 33 \]
\[ k = \frac{33}{7} \]

6. **Conclusion:**
The value of \( k \) for which the system has a non-trivial solution is:
\[ k = \frac{33}{7} \]