**Correct option (D) P, Q and R are non-collinear**

**Explanation :**

It is known that three points (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) are collinear if and only if

Therefore, from R_{3} - R_{1}sin*θ -* R_{2}cos*θ, we have*

This implies that x_{1}y_{2} - x_{2}y_{1 }= 0 or sinθ + cosθ = 1. Since 0 < θ < π/4, sinθ + cosθ *≠* 1. Therefore, x_{1}y_{2} x_{2}y_{1} = 0.

This implies that

-sin(β - α)sin β + cosβcos(β - α) = 0

cos(β - α + β) = 0

Hence,2β - α = π/2 which is impossible because 0 < α and β < π/4. Therefore

2β - α *≠ *π/2

Thus x_{1}y_{2} - x_{2}y_{1} *≠* 0. Hence, the points P, Q and R are non-collinear.