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 set up the determinant of the coefficients of the equations and solve for when this determinant equals zero.

### 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: Set up the determinant
The system of equations can be represented in matrix form as follows:
\[ \begin{vmatrix} \sin 3\theta & -2 & 3 \\ \cos 2\theta & 8 & -7 \\ 2 & 14 & -11 \end{vmatrix} = 0 \]

### Step 2: Calculate the determinant
We will calculate the determinant using the formula for a \(3 \times 3\) determinant:
\[ D = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where the matrix is:
\[ \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} \]

For our matrix, we have:
- \(a = \sin 3\theta\), \(b = -2\), \(c = 3\)
- \(d = \cos 2\theta\), \(e = 8\), \(f = -7\)
- \(g = 2\), \(h = 14\), \(i = -11\)

Calculating the determinant:
\[ D = \sin 3\theta \cdot (8 \cdot (-11) - (-7) \cdot 14) - (-2) \cdot (\cos 2\theta \cdot (-11) - (-7) \cdot 2) + 3 \cdot (\cos 2\theta \cdot 14 - 8 \cdot 2) \]

Calculating each term:
1. \(8 \cdot (-11) = -88\)
2. \((-7) \cdot 14 = -98\)
3. Thus, \(8 \cdot (-11) - (-7) \cdot 14 = -88 + 98 = 10\)

Now substituting back:
\[ D = \sin 3\theta \cdot 10 + 2 \cdot (\cos 2\theta \cdot 11 + 14) + 3 \cdot (14 \cos 2\theta - 16) \]

### Step 3: Simplifying the determinant
Expanding this gives:
\[ D = 10 \sin 3\theta + 2 (11 \cos 2\theta + 14) + 3 (14 \cos 2\theta - 16) \]
\[ = 10 \sin 3\theta + 22 \cos 2\theta + 28 + 42 \cos 2\theta - 48 \]
\[ = 10 \sin 3\theta + 64 \cos 2\theta - 20 \]

### Step 4: Set the determinant to zero
For the system to have a non-trivial solution, we set the determinant to zero:
\[ 10 \sin 3\theta + 64 \cos 2\theta - 20 = 0 \]
\[ 10 \sin 3\theta + 64 \cos 2\theta = 20 \]

### Step 5: Solve for \(\theta\)
We can express \(\sin 3\theta\) and \(\cos 2\theta\) in terms of \(\sin \theta\):
- \(\sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta\)
- \(\cos 2\theta = 1 - 2 \sin^2 \theta\)

Substituting these into the equation gives:
\[ 10(3 \sin \theta - 4 \sin^3 \theta) + 64(1 - 2 \sin^2 \theta) = 20 \]
This simplifies to:
\[ 30 \sin \theta - 40 \sin^3 \theta + 64 - 128 \sin^2 \theta = 20 \]
\[ 30 \sin \theta - 40 \sin^3 \theta - 128 \sin^2 \theta + 44 = 0 \]

### Step 6: Factor and solve the polynomial
Let \(x = \sin \theta\):
\[ -40x^3 - 128x^2 + 30x + 44 = 0 \]

Using the quadratic formula or factoring techniques, we can find the roots.

### Final Step: Find values of \(\theta\)
The possible values of \(x\) (i.e., \(\sin \theta\)) lead to:
1. \(x = -\frac{1}{2}\) gives \(\theta = -\frac{\pi}{6} + n\pi\)
2. \(x = 0\) gives \(\theta = n\pi\)

Thus, the values of \(\theta\) for which the system has a non-trivial solution are:
\[ \theta = n\pi \quad \text{and} \quad \theta = n\pi - \frac{\pi}{6} \]