1. Key Concepts and Formulas

• Non-trivial Solutions for Homogeneous Systems: A system of linear equations $Ax=0$ (where $A$ is the coefficient matrix and $x$ is the variable vector) is called a homogeneous system. It always has a trivial solution ($x=0$). For a homogeneous system to have non-trivial solutions (i.e., at least one variable is non-zero), the determinant of its coefficient matrix must be equal to zero.
• Determinant of a $3 \times 3$ Matrix: For a matrix $A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$, its determinant is given by $\det(A) = a(ei - fh) - b(di - fg) + c(dh - eg)$. Alternatively, row/column operations can be used to simplify the determinant calculation.
• Solving Trigonometric Equations: To find all solutions for $\tan \theta = k$ in a given interval, first find the principal value $\alpha$ (usually in $(-\pi/2, \pi/2)$). The general solution is $\theta = n\pi + \alpha$, where $n$ is an integer. Then, substitute integer values for $n$ to find all solutions within the specified interval.

2. Step-by-Step Solution

Step 1: Formulate the Coefficient Matrix and Apply Non-trivial Solution Condition

The given system of linear equations is: $x+y+\sqrt{3} z=0$ $-x+(\tan \theta) y+\sqrt{7} z=0$ $x+y+(\tan \theta) z=0$ This is a homogeneous system. For it to have non-trivial solutions, the determinant of its coefficient matrix must be zero. The coefficient matrix $A$ is: $A = \begin{pmatrix} 1 & 1 & \sqrt{3} \\ -1 & \tan \theta & \sqrt{7} \\ 1 & 1 & \tan \theta \end{pmatrix}$ We set $\det(A) = 0$.

Step 2: Calculate the Determinant

We can simplify the determinant calculation by performing a column operation. Notice that the first and third rows have identical elements in the first two columns. Let $C_1, C_2, C_3$ be the columns of the matrix. We perform the operation $C_2 \to C_2 - C_1$: $\det(A) = \left|\begin{array}{ccc} 1 & 1-1 & \sqrt{3} \\ -1 & \tan \theta - (-1) & \sqrt{7} \\ 1 & 1-1 & \tan \theta \end{array}\right| = \left|\begin{array}{ccc} 1 & 0 & \sqrt{3} \\ -1 & \tan \theta + 1 & \sqrt{7} \\ 1 & 0 & \tan \theta \end{array}\right|$ Now, expand the determinant along the second column, as it contains two zeros: $\det(A) = 0 \cdot (\dots) - (\tan \theta + 1) \cdot \left|\begin{array}{cc} 1 & \sqrt{3} \\ 1 & \tan \theta \end{array}\right| + 0 \cdot (\dots) = 0$ $-(\tan \theta + 1)(1 \cdot \tan \theta - 1 \cdot \sqrt{3}) = 0$ $-(\tan \theta + 1)(\tan \theta - \sqrt{3}) = 0$

Step 3: Solve the Trigonometric Equation for $\tan \theta$

From the simplified determinant equation, we have two possibilities: $\tan \theta + 1 = 0 \quad \text{or} \quad \tan \theta - \sqrt{3} = 0$ This gives us: $\tan \theta = -1 \quad \text{or} \quad \tan \theta = \sqrt{3}$

Step 4: Find all values of $\theta$ in the interval $[-\pi, \pi]$

We need to find all $\theta \in [-\pi, \pi]$ that satisfy these conditions.

• Case 1: $\tan \theta = \sqrt{3}$ The principal value is $\theta_0 = \frac{\pi}{3}$. The general solution is $\theta = n\pi + \frac{\pi}{3}$, where $n \in \mathbb{Z}$. For $\theta \in [-\pi, \pi]$:

  • If $n=0$, $\theta = \frac{\pi}{3}$.
  • If $n=-1$, $\theta = -\pi + \frac{\pi}{3} = -\frac{2\pi}{3}$. (For $n=1$, $\theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$, which is outside the interval). So, the solutions for $\tan \theta = \sqrt{3}$ are $\left\{\frac{\pi}{3}, -\frac{2\pi}{3}\right\}$.

• Case 2: $\tan \theta = -1$ The principal value is $\theta_0 = -\frac{\pi}{4}$ (or $\frac{3\pi}{4}$). The general solution is $\theta = n\pi - \frac{\pi}{4}$, where $n \in \mathbb{Z}$. For $\theta \in [-\pi, \pi]$:

  • If $n=0$, $\theta = -\frac{\pi}{4}$.
  • If $n=1$, $\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$. (For $n=-1$, $\theta = -\pi - \frac{\pi}{4} = -\frac{5\pi}{4}$, which is outside the interval). So, the solutions for $\tan \theta = -1$ are $\left\{-\frac{\pi}{4}, \frac{3\pi}{4}\right\}$.

Combining these, the set $S$ of all values of $\theta$ is: $S = \left\{\frac{\pi}{3}, -\frac{2\pi}{3}, -\frac{\pi}{4}, \frac{3\pi}{4}\right\}$

Step 5: Calculate the Sum of $\theta$ values

Now, we sum all the values in the set $S$: $\sum_{\theta \in S} \theta = \frac{\pi}{3} - \frac{2\pi}{3} - \frac{\pi}{4} + \frac{3\pi}{4}$ Group terms with common denominators: $= \left(\frac{\pi}{3} - \frac{2\pi}{3}\right) + \left(-\frac{\pi}{4} + \frac{3\pi}{4}\right)$ $= \left(-\frac{\pi}{3}\right) + \left(\frac{2\pi}{4}\right)$ $= -\frac{\pi}{3} + \frac{\pi}{2}$ Find a common denominator (6): $= \frac{-2\pi + 3\pi}{6} = \frac{\pi}{6}$

Step 6: Calculate the Final Expression

Finally, we substitute the sum into the given expression $\frac{120}{\pi} \sum_{\theta \in S} \theta$: $\frac{120}{\pi} \left(\frac{\pi}{6}\right) = \frac{120}{6} = 20$

3. Common Mistakes & Tips

• Determinant Calculation Errors: Be meticulous when expanding determinants. Using row/column operations to introduce zeros can significantly reduce the chance of errors.
• Missing Solutions in Interval: Always remember that trigonometric functions are periodic. After finding principal values, use the general solution form ($\theta = n\pi + \alpha$ for $\tan \theta$) and check all integer values of $n$ to find solutions within the specified interval $[-\pi, \pi]$. Do not forget negative solutions.
• Undefined Tangent: Ensure that the values of $\theta$ obtained do not make $\tan \theta$ undefined (i.e., $\theta \neq \frac{\pi}{2} + k\pi$). In this case, all obtained values are valid.

4. Summary

The problem required finding values of $\theta$ for which a homogeneous system of linear equations has non-trivial solutions. This condition translates to setting the determinant of the coefficient matrix to zero. Solving the resulting determinant equation led to a quadratic equation in $\tan \theta$, yielding $\tan \theta = \sqrt{3}$ and $\tan \theta = -1$. We then found all corresponding $\theta$ values within the interval $[-\pi, \pi]$. Summing these values gave $\frac{\pi}{6}$, and substituting this into the final expression $\frac{120}{\pi} \sum_{\theta \in S} \theta$ resulted in 20.

5. Final Answer The final answer is $\boxed{40}$. This corresponds to option (A).