To solve this question, we need to understand the concept of Coulomb's Law and the principle of superposition. Coulomb's Law states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. The principle of superposition states that the total force on a charged particle due to multiple other charged particles is the vector sum of the individual forces.

Let's analyze the given situation step by step:

1. The first particle is at the origin, represented as 'o', and has a charge 'q'.

2. The second particle is at a distance 'd' from the first particle, represented as 'p', and also has a charge 'q'.

3. The third particle is at a distance '2d' from the first particle, represented as 'q', and has a charge '2q'.

Now, let's calculate the net force on each particle:

a) Particle at the origin ('o'):

Since there are no other charged particles around, the net force on particle 'o' is zero.

b) Particle at distance 'd' ('p'):

There are two charged particles affecting particle 'p' - one at the origin and the other at distance '2d'.

- The force between 'o' and 'p' is given by Coulomb's Law: F = k*q*q/d^2, where k is the electrostatic constant.
- The force between 'o' and 'p' is attractive, so it is directed towards 'o'.
- The magnitude of the force is F1 = k*q*q/d^2.

- The force between 'p' and 'q' is given by Coulomb's Law: F = k*q*(2q)/(2d)^2.
- The force between 'p' and 'q' is repulsive, so it is directed away from 'q'.
- The magnitude of the force is F2 = k*q*(2q)/(2d)^2.

According to the principle of superposition, the net force on particle 'p' is the vector sum of the individual forces: F_net = F1 + F2.

c) Particle at distance '2d' ('q'):

There is only one charged particle affecting particle 'q' - the one at distance 'd'.

- The force between 'q' and 'p' is given by Coulomb's Law: F = k*(2q)*q/(2d)^2.
- The force between 'q' and 'p' is attractive, so it is directed towards 'q'.
- The magnitude of the force is F3 = k*(2q)*q/(2d)^2.

Therefore, the net force on particle 'q' is the vector sum of the individual force: F_net = F3.

d) Particle at distance '3d' ('r'):

Since there are no other charged particles around, the net force on particle 'r' is zero.

Conclusion:

Based on the analysis, the correct answer is option 'C' (Particle at distance 'd' experiences no net force). The net force on the particle at distance 'd' is zero because the attractive force from the particle at the origin is canceled out by the repulsive force from the particle at distance '2d'.