If the system of linear equations `x+ky+3z=0, 3x+ky-2z=0, 2x+4y-3z=0` has a non-zero solution `(x,y,z)` then `(xz)/(y^2)` is equal to

Explanation

To solve the system of linear equations given by:

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

we need to find the value of \( \frac{xz}{y^2} \) when there is a non-zero solution for \( (x, y, z) \).

### Step 1: Formulate the system in matrix form

We can express the system of equations in matrix form as follows:

\[ \begin{bmatrix} 1 & k & 3 \\ 3 & k & -2 \\ 2 & 4 & -3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

### Step 2: Calculate the determinant

For the system to have a non-zero solution, the determinant of the coefficient matrix must be zero:

\[ \Delta = \begin{vmatrix} 1 & k & 3 \\ 3 & k & -2 \\ 2 & 4 & -3 \end{vmatrix} \]

Calculating the determinant:

\[ \Delta = 1 \begin{vmatrix} k & -2 \\ 4 & -3 \end{vmatrix} - k \begin{vmatrix} 3 & -2 \\ 2 & -3 \end{vmatrix} + 3 \begin{vmatrix} 3 & k \\ 2 & 4 \end{vmatrix} \]

Calculating the minors:

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

2. \( \begin{vmatrix} 3 & -2 \\ 2 & -3 \end{vmatrix} = 3(-3) - (-2)(2) = -9 + 4 = -5 \)

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

Putting it all together:

\[ \Delta = 1(-3k + 8) - k(-5) + 3(12 - 2k) \]
\[ = -3k + 8 + 5k + 36 - 6k \]
\[ = (-3k + 5k - 6k) + (8 + 36) = -4k + 44 \]

Setting the determinant to zero for a non-zero solution:

\[ -4k + 44 = 0 \implies 4k = 44 \implies k = 11 \]

### Step 3: Substitute \( k \) back into the equations

Now substituting \( k = 11 \) into the original equations:

1. \( x + 11y + 3z = 0 \)
2. \( 3x + 11y - 2z = 0 \)
3. \( 2x + 4y - 3z = 0 \)

### Step 4: Solve the equations

From the first equation:

\[ z = -\frac{x + 11y}{3} \]

Substituting \( z \) into the second equation:

\[ 3x + 11y - 2\left(-\frac{x + 11y}{3}\right) = 0 \]
\[ 3x + 11y + \frac{2x + 22y}{3} = 0 \]
Multiplying through by 3 to eliminate the fraction:

\[ 9x + 33y + 2x + 22y = 0 \implies 11x + 55y = 0 \implies x = -5y \]

Now substituting \( x = -5y \) into the expression for \( z \):

\[ z = -\frac{-5y + 11y}{3} = -\frac{6y}{3} = -2y \]

### Step 5: Calculate \( \frac{xz}{y^2} \)

Now we have:

- \( x = -5y \)
- \( z = -2y \)

Calculating \( \frac{xz}{y^2} \):

\[ \frac{xz}{y^2} = \frac{(-5y)(-2y)}{y^2} = \frac{10y^2}{y^2} = 10 \]

### Final Answer

Thus, the value of \( \frac{xz}{y^2} \) is:

\[ \boxed{10} \]