System of equations `x+2y+z=0, 2x+3y-z=0` and `(tantheta)x+y-3z=0` has non-trival solution then number of value (s) of `theta epsilon(-pi,pi)` is equal to

Explanation

To solve the given system of equations and determine the number of values of \(\theta\) in the interval \((- \pi, \pi)\) for which the system has a non-trivial solution, we follow these steps:

### Step 1: Write the system of equations in matrix form
The equations given are:
1. \(x + 2y + z = 0\)
2. \(2x + 3y - z = 0\)
3. \((\tan \theta)x + y - 3z = 0\)

We can represent this system in matrix form as:
\[ \begin{bmatrix} 1 & 2 & 1 \\ 2 & 3 & -1 \\ \tan \theta & 1 & -3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

### Step 2: Set the determinant of the coefficient matrix to zero
For the system to have a non-trivial solution, the determinant of the coefficient matrix must be zero:
\[ \text{det} \begin{bmatrix} 1 & 2 & 1 \\ 2 & 3 & -1 \\ \tan \theta & 1 & -3 \end{bmatrix} = 0 \]

### Step 3: Calculate the determinant
We can calculate the determinant using the formula for a \(3 \times 3\) matrix:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where \(A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}\).

Substituting the values:
\[ = 1 \cdot (3 \cdot (-3) - (-1) \cdot 1) - 2 \cdot (2 \cdot (-3) - (-1) \cdot \tan \theta) + 1 \cdot (2 \cdot 1 - 3 \cdot \tan \theta) \]

Calculating each term:
1. \(1 \cdot (-9 + 1) = 1 \cdot (-8) = -8\)
2. \(-2 \cdot (-6 + \tan \theta) = 12 - 2\tan \theta\)
3. \(1 \cdot (2 - 3\tan \theta) = 2 - 3\tan \theta\)

Combining these gives:
\[ -8 + 12 - 2\tan \theta + 2 - 3\tan \theta = 0 \]
Simplifying:
\[ 6 - 5\tan \theta = 0 \]

### Step 4: Solve for \(\tan \theta\)
Setting the equation to zero:
\[ -5\tan \theta = -6 \implies \tan \theta = \frac{6}{5} \]

### Step 5: Determine the values of \(\theta\)
Now we need to find the values of \(\theta\) in the interval \((- \pi, \pi)\) for which \(\tan \theta = \frac{6}{5}\).

The tangent function is periodic with a period of \(\pi\). Therefore, the general solutions for \(\tan \theta = \frac{6}{5}\) are:
\[ \theta = \tan^{-1}\left(\frac{6}{5}\right) + n\pi \quad (n \in \mathbb{Z}) \]

### Step 6: Find specific solutions in \((- \pi, \pi)\)
1. For \(n = 0\):
\[ \theta_1 = \tan^{-1}\left(\frac{6}{5}\right) \]
2. For \(n = -1\):
\[ \theta_2 = \tan^{-1}\left(\frac{6}{5}\right) - \pi \]
3. For \(n = 1\):
\[ \theta_3 = \tan^{-1}\left(\frac{6}{5}\right) + \pi \quad \text{(not in the interval)} \]

Thus, the two values of \(\theta\) in the interval \((- \pi, \pi)\) are:
- \(\theta_1 = \tan^{-1}\left(\frac{6}{5}\right)\)
- \(\theta_2 = \tan^{-1}\left(\frac{6}{5}\right) - \pi\)

### Conclusion
The number of values of \(\theta\) in the interval \((- \pi, \pi)\) is **2**.