Refer to Fig. `O` is the centriod of triangle `ABC`, where
`OA = (2)/(3) AD = (2)/(3) (AB sin 60^(@))`
`= (2)/(3) xx a xx (sqrt(3))/(2) = (a)/(sqrt(3))`
Thus, `OA = OB = OC = (a)/(sqrt(3))`
The gravitational intensity at `O` due to mass `m`
at `A` is, `I_(A) = (Gm)/((OA)^(2)) = (Gm)/((a//sqrt(3))^(2))` along `OA`.
Similarly the gravitational intensity at `O` due to mass `m` at `B` is,
`I_(B) = (Gm)/((OB)^(2)) = (Gm)/((a//sqrt(3))^(2))` along `OB`.
and gravitational intensity at `O` due to mass `m`
at `C` is,
`I_(C) = (Gm)/((OC)^(2)) = (Gm)/((a//sqrt(3))^(2))` along `OC`.
As `I_(A),I_(B)` and `I_(C)` are equal in magnitude and equally inclined to each other, the resultant gravitational intensity at `O` is zero.
gravitational potential at `O` due to masses at
`A,B` and `C` is
`V = - (Gm)/(OA) + (-(Gm)/(OB)) + (-(Gm)/(OC))`
`= - (3 Gm)/(OA) = (-3 Gm)/(a//sqrt(3)) = (-3 sqrt(3) Gm)/(a)`