C

Step 1: Understanding the initial problem setup. There are two particles, say P1 and P2, moving horizontally with initial velocities:
- Particle P1 has an initial velocity of 3 m/s to the right.
- Particle P2 has an initial velocity of 4 m/s to the left.
Both particles are subject to gravitational acceleration (g) downwards.

Step 2: Write the equations of motion for both particles.
For P1:
- Horizontal position: \( x_1(t) = 3t \)
- Vertical position: \( y_1(t) = -\frac{1}{2}gt^2 \)
For P2:
- Horizontal position: \( x_2(t) = -4t \)
- Vertical position: \( y_2(t) = -\frac{1}{2}gt^2 \)

Step 3: Find the moment when the velocity vectors become mutually perpendicular.
The velocities of the particles at time \( t \) are given as:
- For P1: \( v_{1x} = 3 \) m/s, \( v_{1y} = -gt \)
- For P2: \( v_{2x} = -4 \) m/s, \( v_{2y} = -gt \)

The dot product of the velocity vectors must be zero for the vectors to be perpendicular:
\( v_{1x} v_{2x} + v_{1y} v_{2y} = 0 \)
Substituting the expressions:
\( (3)(-4) + (-gt)(-gt) = 0 \)

Step 4: Simplifying the equation:
\( -12 + g^2 t^2 = 0 \)
This leads to:
\( g^2 t^2 = 12 \)
Therefore:
\( t^2 = \frac{12}{g^2} \)

Step 5: Substitute for the time t that we got into the position equations to find the distances.
The distance traveled by P1 horizontally:
\( d_1 = 3t = 3\sqrt{\frac{12}{g^2}} = \frac{3\sqrt{12}}{g} = \frac{6\sqrt{3}}{g} \)
The distance traveled by P2 is:
\( d_2 = 4t = 4\sqrt{\frac{12}{g^2}} = \frac{4\sqrt{12}}{g} = \frac{8\sqrt{3}}{g} \)

Step 6: The total distance between the two particles when their velocities are perpendicular is:
\( D = d_1 + d_2 = \frac{6\sqrt{3}}{g} + \frac{8\sqrt{3}}{g} = \frac{14\sqrt{3}}{g} \)

Assuming g = 9.8 m/s² (for calculations), substitute the value to find the distance.
\( D = \frac{14\sqrt{3}}{9.8} \approx 2.55 \) m.
The closest answer choice, based on possible options, is C.