**Solution:**

To find the acceleration of the particle when the third force is removed, we need to consider the net force acting on the particle.

**1. Net Force:**

The net force on a particle can be found by adding the vector sum of all the forces acting on it. In this case, we have three forces R1, R2, and R3.

Net force (F_net) = R1 + R2 + R3

**2. Equilibrium Condition:**

Initially, the particle is in equilibrium, which means the net force acting on it is zero. This can be expressed as:

F_net = 0

Therefore, when the forces R1 and R2 are acting on the particle, R3 must be such that it cancels out the vector sum of R1 and R2.

**3. Perpendicular Forces:**

Given that R1 and R2 are perpendicular to each other, we can express them as:

R1 = R1_xi + R1_yj
R2 = R2_xi + R2_yj

Here, R1_x and R1_y represent the x and y components of force R1, respectively. Similarly, R2_x and R2_y represent the x and y components of force R2.

**4. Removing R3:**

When the third force R3 is removed, we can rewrite the net force equation as:

F_net = R1 + R2

Since the particle is initially in equilibrium, the net force is zero. Therefore, we can write:

R1 + R2 = 0

**5. Acceleration:**

The acceleration of the particle can be found using Newton's second law, which states that the acceleration (a) is equal to the net force (F_net) divided by the mass (m) of the particle.

a = F_net / m

In this case, the net force is zero (R1 + R2 = 0) and the mass is given as m. Therefore, the acceleration of the particle is zero.

**Conclusion:**

When the third force R3 is removed, the particle remains in equilibrium and does not experience any acceleration. This is because the vector sum of the forces R1 and R2 is already canceled out by R3 in the initial equilibrium state. Thus, the particle will continue to remain at rest or in a state of uniform motion in a straight line.