let particle are at A,B,C...

force on A due to B is F & force on A due to C is also F...

F = Gm²/a²

net force  = (F²+F²+2F²cos@)^1/2                                 (by vector method)

angle bw them is @ = 60

so net force = F_n = 3^1/2F       ...........1

due to this force ,each particle circulates aroung their center of mass & center of mass lies at the center of equilateral triangle ......

F_c = F_n = 3^1/2Gm²/a²   

mv²/r = 3^1/2Gm²/a²

  v = {3^1/2 Gm/a²r}^1/2 

 now r = a/3^1/2        

 now v becomes

 v = {3Gm/a³}^1/2                                             ans1

time period = 2pir/v

                =2pi(a/3^1/2){3Gm/a³)^1/2

               = 2pi(gm/a)^1/2                             ans2

Approved