Step 1: Using the formula for intensity in a Young's double slit experiment.
The intensity at any point in Young's double slit experiment is given by: \[ I_{\text{res}} = 4I_0 \cos^2 \left( \frac{\theta}{2} \right) \] Where \( I_0 \) is the maximum intensity, and \( \theta \) is the phase difference. For the first point \( P \), where \( \theta = \frac{\pi}{3} \): \[ I_1 = 4I_0 \cos^2 \left( \frac{\pi}{6} \right) = 4I_0 \left( \frac{\sqrt{3}}{2} \right)^2 = 4I_0 \times \frac{3}{4} = 3I_0 \] For the second point \( Q \), where \( \theta = \frac{\pi}{2} \): \[ I_2 = 4I_0 \cos^2 \left( \frac{\pi}{4} \right) = 4I_0 \left( \frac{1}{\sqrt{2}} \right)^2 = 4I_0 \times \frac{1}{2} = 2I_0 \] Thus, the ratio of intensities is: \[ \frac{I_1}{I_2} = \frac{3I_0}{2I_0} = \frac{3}{2} \]

Let \( a \in \mathbb{R} \) and \( A \) be a matrix of order \( 3 \times 3 \) such that \( \det(A) = -4 \) and \[ A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix} \] where \( I \) is the identity matrix of order \( 3 \times 3 \).
If \( \det\left( (a + 1) \cdot \text{adj}\left( (a - 1) A \right) \right) \) is \( 2^m 3^n \), \( m, n \in \{ 0, 1, 2, \dots, 20 \} \), then \( m + n \) is equal to: