The determinant is given by:
\[ D = \left| \begin{matrix} 1 & 1 & \sqrt{3} \\ -1 & \tan \theta & \sqrt{7} \\ 1 & 1 & \tan \theta \end{matrix} \right| = 0 \]
Simplifying the determinant expression: \[ \tan^2 \theta - (\sqrt{3} - 1) - \sqrt{3} = 0 \]
Solving for \( \tan \theta \): \[ \tan \theta = \sqrt{3}, -1 \]
Therefore, the possible values for \( \theta \) are: \[ \theta = \left\{ \frac{\pi}{3}, -\frac{2\pi}{3}, -\frac{\pi}{4}, \frac{3\pi}{4} \right\} \]
Now, calculating the sum of angles: \[ \frac{120}{\pi} \left( \sum \theta \right) = \frac{120}{\pi} \times \frac{\pi}{6} = 20 \quad (\text{Option 2}) \]
In the following \(p\text{–}V\) diagram, the equation of state along the curved path is given by \[ (V-2)^2 = 4ap, \] where \(a\) is a constant. The total work done in the closed path is: 
Let \( ABC \) be a triangle. Consider four points \( p_1, p_2, p_3, p_4 \) on the side \( AB \), five points \( p_5, p_6, p_7, p_8, p_9 \) on the side \( BC \), and four points \( p_{10}, p_{11}, p_{12}, p_{13} \) on the side \( AC \). None of these points is a vertex of the triangle \( ABC \). Then the total number of pentagons that can be formed by taking all the vertices from the points \( p_1, p_2, \ldots, p_{13} \) is ___________.
Consider the following two reactions A and B: 
The numerical value of [molar mass of $x$ + molar mass of $y$] is ___.