Three point particles $P, Q, R$ move in a circle of radius $\text{'}r\text{'}$ with different but constant speeds. They start moving at $t=0$ from their initial positions as shown in the figure. The angular velocities (in $\raisebox{1ex}{$rad$}\!\left/ \!\raisebox{-1ex}{$sec$}\right.$) of $P, Q$ and $R$ are $5\mathrm{\pi }, 2\mathrm{\pi }$ and $3\mathrm{\pi }$ respectively, in the same sense. 

$1st$ check when $Q and R$ will meet , $R$ is $\mathrm{\pi } rad/s$ faster then $Q$. 

For the first time they will meet 

$t=\frac{{\displaystyle \raisebox{1ex}{$3\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{\mathrm{\pi }}$ (Relative angle between them =$\frac{3\mathrm{\pi }}{2})$

$t=\frac{3}{2} sec.$ 

Now they will meet again after $2\mathrm{\pi } rad$ 

So time is $\frac{2\mathrm{\pi }}{\mathrm{\pi }}=2 sec.$ 

For $Q and R , t=\frac{3}{2} s, (\frac{3}{2}+2) s,( \frac{3}{2}+2+2) s, .....$ 

Similarly for $P$ and $R$ 

$t=\frac{1}{2}s, (\frac{1}{2}+1)s,( \frac{1}{2}+1+1)s$ 

So $3$ will meet at $\frac{3}{2}s$ for the $1st$ time. 

 $P and Q$ will meet at

$$\begin{gathered} t=\frac{1}{6} s,( \frac{1}{6}+\frac{2}{3}) s,( \frac{1}{6}+\frac{4}{3})s, ..... \\ =\frac{1}{6}s, \frac{5}{6}s, \frac{9}{6}s .... \end{gathered}$$ 

So P and Q will meet $3$ times in this interval of time ($\frac{3}{2}s).$