Number of values of `theta` lying I [0,100`pi`] for which the system of equations (sin 3`theta`) x-y+z=0, (cos 2`theta`) x+4y +3z=0, 2x+ 7y+7z =0 has non-trivial solution is `"____"`

Explanation

To find the number of values of \( \theta \) in the interval \([0, 100\pi]\) for which the system of equations has a non-trivial solution, we need to analyze the determinant of the coefficient matrix of the system. The system of equations is given as:

1. \( \sin(3\theta) x - y + z = 0 \)
2. \( \cos(2\theta) x + 4y + 3z = 0 \)
3. \( 2x + 7y + 7z = 0 \)

### Step 1: Form the Coefficient Matrix and Set Up the Determinant
The coefficient matrix of the system can be written as:

\[ \begin{bmatrix} \sin(3\theta) & -1 & 1 \\ \cos(2\theta) & 4 & 3 \\ 2 & 7 & 7 \end{bmatrix} \]

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

\[ \Delta = \begin{vmatrix} \sin(3\theta) & -1 & 1 \\ \cos(2\theta) & 4 & 3 \\ 2 & 7 & 7 \end{vmatrix} = 0 \]

### Step 2: Calculate the Determinant
Calculating the determinant using the formula for a \(3 \times 3\) matrix:

\[ \Delta = \sin(3\theta) \begin{vmatrix} 4 & 3 \\ 7 & 7 \end{vmatrix} - (-1) \begin{vmatrix} \cos(2\theta) & 3 \\ 2 & 7 \end{vmatrix} + 1 \begin{vmatrix} \cos(2\theta) & 4 \\ 2 & 7 \end{vmatrix} \]

Calculating the minors:

1. \( \begin{vmatrix} 4 & 3 \\ 7 & 7 \end{vmatrix} = 4 \cdot 7 - 3 \cdot 7 = 28 - 21 = 7 \)
2. \( \begin{vmatrix} \cos(2\theta) & 3 \\ 2 & 7 \end{vmatrix} = \cos(2\theta) \cdot 7 - 3 \cdot 2 = 7\cos(2\theta) - 6 \)
3. \( \begin{vmatrix} \cos(2\theta) & 4 \\ 2 & 7 \end{vmatrix} = \cos(2\theta) \cdot 7 - 4 \cdot 2 = 7\cos(2\theta) - 8 \)

Substituting back into the determinant:

\[ \Delta = \sin(3\theta) \cdot 7 + (7\cos(2\theta) - 6) + (7\cos(2\theta) - 8) \]

Combining terms:

\[ \Delta = 7\sin(3\theta) + 14\cos(2\theta) - 14 = 0 \]

### Step 3: Simplify the Equation
Dividing the entire equation by 7 gives:

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

### Step 4: Use Trigonometric Identities
Using the identities:

- \( \sin(3\theta) = 3\sin(\theta) - 4\sin^3(\theta) \)
- \( \cos(2\theta) = 1 - 2\cos^2(\theta) \)

Substituting these into the equation:

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

This simplifies to:

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

### Step 5: Factor the Equation
Rearranging gives:

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

Factoring out \( \sin(\theta) \):

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

### Step 6: Solve for \( \theta \)
1. **First Factor**: \( \sin(\theta) = 0 \)
- Solutions in \([0, 100\pi]\): \( \theta = n\pi \) where \( n = 0, 1, 2, \ldots, 100 \) (Total: 101 solutions).

2. **Second Factor**: \( 4\sin^2(\theta) + 4\sin(\theta) - 3 = 0 \)
- Using the quadratic formula:
\[ \sin(\theta) = \frac{-4 \pm \sqrt{16 + 48}}{8} = \frac{-4 \pm 8}{8} \]
- Solutions: \( \sin(\theta) = \frac{1}{2} \) (valid) and \( \sin(\theta) = -\frac{3}{2} \) (not valid).
- For \( \sin(\theta) = \frac{1}{2} \):
- Solutions in \([0, 100\pi]\): \( \theta = \frac{\pi}{6} + 2k\pi \) and \( \theta = \frac{5\pi}{6} + 2k\pi \) for \( k = 0, 1, 2, \ldots, 99 \) (Total: 100 solutions).

### Final Count of Solutions
Adding the solutions from both factors gives:

\[ 101 + 100 = 201 \]

### Answer
The number of values of \( \theta \) lying in \([0, 100\pi]\) for which the system of equations has a non-trivial solution is **201**.