the value of `theta` for which the system of equations `(sin 3theta)x-2y+3z=0, (cos 2theta)x+8y-7z=0`
and `2x+14y-11z=0` has a non - trivial solution, is (here, `n in Z`)

Options

A. `npi`
B. `npi+(-1)^(n)pi//3`
C. `npi+(-1)^(n)pi//2`
D. None of these

Answer

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

1. \( \sin(3\theta)x - 2y + 3z = 0 \)
2. \( \cos(2\theta)x + 8y - 7z = 0 \)
3. \( 2x + 14y - 11z = 0 \)

### Step 1: Form the Coefficient Matrix
We can represent the system of equations in matrix form as follows:

\[ \begin{bmatrix} \sin(3\theta) & -2 & 3 \\ \cos(2\theta) & 8 & -7 \\ 2 & 14 & -11 \end{bmatrix} \]

### Step 2: Set the Determinant to Zero
For the system to have a non-trivial solution, the determinant of the coefficient matrix must be equal to zero:

\[ D = \begin{vmatrix} \sin(3\theta) & -2 & 3 \\ \cos(2\theta) & 8 & -7 \\ 2 & 14 & -11 \end{vmatrix} = 0 \]

### Step 3: Calculate the Determinant
Using the determinant formula for a 3x3 matrix, we have:

\[ D = \sin(3\theta) \begin{vmatrix} 8 & -7 \\ 14 & -11 \end{vmatrix} + 2 \begin{vmatrix} \cos(2\theta) & -7 \\ 2 & -11 \end{vmatrix} + 3 \begin{vmatrix} \cos(2\theta) & 8 \\ 2 & 14 \end{vmatrix} \]

Calculating the minors:

1. \( \begin{vmatrix} 8 & -7 \\ 14 & -11 \end{vmatrix} = (8)(-11) - (-7)(14) = -88 + 98 = 10 \)
2. \( \begin{vmatrix} \cos(2\theta) & -7 \\ 2 & -11 \end{vmatrix} = \cos(2\theta)(-11) - (-7)(2) = -11\cos(2\theta) + 14 \)
3. \( \begin{vmatrix} \cos(2\theta) & 8 \\ 2 & 14 \end{vmatrix} = \cos(2\theta)(14) - (8)(2) = 14\cos(2\theta) - 16 \)

Substituting these back into the determinant:

\[ D = \sin(3\theta)(10) + 2(-11\cos(2\theta) + 14) + 3(14\cos(2\theta) - 16) \]

### Step 4: Simplify the Determinant
Now we simplify:

\[ D = 10\sin(3\theta) - 22\cos(2\theta) + 28 + 42\cos(2\theta) - 48 \]
\[ D = 10\sin(3\theta) + 20\cos(2\theta) - 20 \]

### Step 5: Set the Determinant to Zero
Setting \( D = 0 \):

\[ 10\sin(3\theta) + 20\cos(2\theta) - 20 = 0 \]
\[ \sin(3\theta) + 2\cos(2\theta) - 2 = 0 \]

### Step 6: Solve for \( \theta \)
Using the identity \( \sin(3\theta) = 3\sin(\theta) - 4\sin^3(\theta) \):

\[ 3\sin(\theta) - 4\sin^3(\theta) + 2(1 - 2\sin^2(\theta)) - 2 = 0 \]
\[ 3\sin(\theta) - 4\sin^3(\theta) - 4\sin^2(\theta) = 0 \]
Factoring out \( \sin(\theta) \):

\[ \sin(\theta)(-4\sin^2(\theta) + 3\sin(\theta) - 4) = 0 \]

### Step 7: Find Solutions
1. \( \sin(\theta) = 0 \) gives \( \theta = n\pi \) where \( n \in \mathbb{Z} \).
2. The quadratic \( -4\sin^2(\theta) + 3\sin(\theta) - 4 = 0 \) can be solved using the quadratic formula.

However, since \( \sin(\theta) \) cannot be greater than 1 or less than -1, we discard any solutions that lead to \( \sin(\theta) = -\frac{3}{8} \) or similar invalid values.

### Final Result
The valid solutions for \( \theta \) are:

\[ \theta = n\pi \quad \text{(where \( n \in \mathbb{Z} \))} \]