The relative velocity of the boat with respect to the water is given by subtracting the velocity of the water from the velocity of the boat.

\[
\vec{v}_{\text{bw}} = \vec{v}_{\text{boat}} - \vec{v}_{\text{water}}
\]

Given:
\[
\vec{v}_{\text{boat}} = 3\hat{i} + 4\hat{j}, \quad \vec{v}_{\text{water}} = -3\hat{i} - 4\hat{j}
\]

Now, subtract:

\[
\vec{v}_{\text{bw}} = (3\hat{i} + 4\hat{j}) - (-3\hat{i} - 4\hat{j})
\]
\[
\vec{v}_{\text{bw}} = 3\hat{i} + 4\hat{j} + 3\hat{i} + 4\hat{j}
\]
\[
\vec{v}_{\text{bw}} = 6\hat{i} + 8\hat{j}
\]

Thus, the relative velocity of the boat with respect to the water is:

\[ {6\hat{i} + 8\hat{j}} \]

To determine when the four persons \( P, Q, R, \) and \( S \) will meet, consider the following:

Relative Velocity:
- Each person moves with speed \( u \) towards the center of the square.

Configuration:
- As they face each other and move towards the center, their paths converge.

Effective Speed Towards Each Other:
- The effective speed of each person towards the center is \( u \cos 45^\circ = \frac{u}{\sqrt{2}} \) because they move diagonally.

Distance to Center:
- The distance from each person to the center of the square is \( \frac{d}{\sqrt{2}} \).

Time to Meet:
\[
t = \frac{\text{Distance}}{\text{Effective Speed}} = \frac{\frac{d}{\sqrt{2}}}{\frac{u}{\sqrt{2}}} = \frac{d}{u}
\]

Thus, the time after which they will all meet is \( \frac{d}{u} \).

Boat speed in still water \(v_b = 6\text{ km/h}\), river speed \(v_r = 2\text{ km/h}\). Upstream speed \(v_u = v_b - v_r = 4\text{ km/h}\). Downstream speed \(v_d = v_b + v_r = 8\text{ km/h}\). Time upstream \(t_u = 2\text{ km} / 4\text{ km/h} = 0.5\) hours. Time downstream \(t_d = 2\text{ km} / 8\text{ km/h} = 0.25\) hours. Total time = \(0.5 + 0.25 = 0.75\) hours = \(45\) minutes.

Relative velocity is given by \(\vec{v}_{AB} = \vec{v}_A - \vec{v}_B = (3\hat{i}+4\hat{j}) - (7\hat{i}-3\hat{j}) = -4\hat{i}+7\hat{j}\) m/s. Therefore, option D is correct.

For minimum crossing time, the swimmer must head perpendicular to the river bank. The net ground velocity is the vector sum of swimmer's velocity and river velocity: \(v_g = \sqrt{v_{\text{sw}}^2 + v_r^2} = \sqrt{8^2 + 4^2} = \sqrt{80} \approx 8.94\text{ m/s} \approx 9\text{ m/s}\).

The time to cross a river is \(t = \frac{d}{v \sin \theta}\), where \(theta\) is the angle with the river flow. For \(t\) to be minimum, \(sin \theta\) must be maximum, which occurs at \(\theta = 90^\circ = \frac{\pi}{2}\).

Assertion (A): For particles to meet, their relative perpendicular velocity component must be zero, meaning their perpendicular velocities must be equal. Otherwise, they would move apart perpendicular to the line joining them. So (A) is true.

Reason (R): Relative velocity is a vector difference and can be calculated in any direction, not exclusively along the line joining two bodies. So (R) is false.

The time to cross the river depends on the component of velocity perpendicular to the river flow (\(v_{\text{normal}}\)). For simultaneous crossing, \(v_{\text{normal}}\) must be equal for both boats. If total speed relative to river is same, and \(v_{\text{normal}}\) is same, then the magnitude of the parallel component is also same. Different paths result from different directions of the parallel component. Thus, (A) is true and (R) is true and explains (A).