To solve this problem, let's consider the projectile motion of the particle on the inclined plane. We assume that there is no air resistance and neglect any friction between the particle and the inclined plane.

1. Projectile Motion on an Inclined Plane:
When a particle is projected on an inclined plane, its motion can be resolved into two components - one parallel to the inclined plane (horizontal direction) and the other perpendicular to the inclined plane (vertical direction).

The horizontal component of velocity (Vx) remains constant throughout the motion, while the vertical component of velocity (Vy) changes due to the effect of gravity.

2. Angle of Inclination (Θ):
The angle of inclination of the plane is the angle between the inclined plane and the horizontal ground. Let's assume this angle is Θ.

3. Angle of Projection (α):
The angle of projection from the inclined plane is the angle between the initial velocity of the particle and the inclined plane. Let's assume this angle is α.

4. Initial Velocity (V):
The initial velocity of the particle can be resolved into two components:
- Vx = V * cos(α) (horizontal component)
- Vy = V * sin(α) (vertical component)

5. Relation between Θ and α:
When the particle strikes the inclined plane at right angles, it means that the vertical component of velocity becomes zero. Therefore, Vy = 0.

Using the equation Vy = V * sin(α), we can write:
V * sin(α) = 0

Since sin(α) ≠ 0 (as it is not possible for α to be 90 degrees), we can conclude that V = 0.

Now, using the equation Vx = V * cos(α), we have:
V * cos(α) = 0

Since cos(α) ≠ 0 (as it is not possible for α to be 0 or 180 degrees), we can conclude that V = 0.

Therefore, the relation between Θ and α is given by:
2tan(Θ) = cot(α)

Hence, the correct option is (a) 2tan(Θ) = cot(α).