If the system of equations
`x-ky+3z=0,`
`2x+ky-2z=0` and `3x-4y+2z=0` has non - trivial solutions, then the value of `(10y)/(x)` is equal to

Options

A. 3
B. `-(15)/(2)`
C. `(5)/(7)`
D. `-(5)/(7)`

Answer

To solve the given system of equations for non-trivial solutions and find the value of \(\frac{10y}{x}\), we can follow these steps:

### Step 1: Write the System of Equations
The given equations are:
1. \(x - ky + 3z = 0\)
2. \(2x + ky - 2z = 0\)
3. \(3x - 4y + 2z = 0\)

### Step 2: Form the Coefficient Matrix
We need to form the coefficient matrix from the coefficients of \(x\), \(y\), and \(z\):
\[ A = \begin{bmatrix} 1 & -k & 3 \\ 2 & k & -2 \\ 3 & -4 & 2 \end{bmatrix} \]

### Step 3: Set the Determinant Equal to Zero
For the system to have non-trivial solutions, the determinant of the coefficient matrix must be zero:
\[ \text{det}(A) = 0 \]

### Step 4: Calculate the Determinant
Calculating the determinant using the formula for a \(3 \times 3\) matrix:
\[ \text{det}(A) = 1 \cdot (k \cdot 2 - (-2)(-4)) - (-k) \cdot (2 \cdot 2 - (-2)(3)) + 3 \cdot (2 \cdot (-4) - k \cdot 3) \]
Calculating each term:
1. \(1 \cdot (2k - 8)\)
2. \(-(-k) \cdot (4 - (-6)) = k \cdot 10\)
3. \(3 \cdot (-8 - 3k)\)

Putting it all together:
\[ \text{det}(A) = 2k - 8 + 10k - 24 - 24 - 9k = 0 \]
Combining like terms:
\[ (2k + 10k - 9k) - 56 = 0 \implies 3k - 56 = 0 \]

### Step 5: Solve for \(k\)
\[ 3k = 56 \implies k = \frac{56}{3} \]

### Step 6: Substitute \(k\) Back into the Equations
Now we substitute \(k\) back into the first equation:
\[ x - \frac{56}{3}y + 3z = 0 \implies x = \frac{56}{3}y - 3z \]

### Step 7: Express \(z\) in Terms of \(x\) and \(y\)
From the second equation:
\[ 2x + \frac{56}{3}y - 2z = 0 \implies 2z = 2x + \frac{56}{3}y \implies z = x + \frac{28}{3}y \]

### Step 8: Substitute \(z\) into the First Equation
Substituting \(z\) into the first equation:
\[ x = \frac{56}{3}y - 3\left(x + \frac{28}{3}y\right) \]
Simplifying:
\[ x = \frac{56}{3}y - 3x - 28y \]
\[ x + 3x = \frac{56}{3}y - 28y \implies 4x = \frac{56}{3}y - \frac{84}{3}y \]
\[ 4x = \frac{-28}{3}y \implies x = -\frac{7}{3}y \]

### Step 9: Find \(\frac{10y}{x}\)
Now we can find \(\frac{10y}{x}\):
\[ \frac{10y}{x} = \frac{10y}{-\frac{7}{3}y} = \frac{10 \cdot 3}{-7} = -\frac{30}{7} \]

### Final Answer
Thus, the value of \(\frac{10y}{x}\) is:
\[ \boxed{-\frac{30}{7}} \]