Three particles, each of mass ‘m’ are placed at the three corners of an equilateral triangle of side L. The force exerted by the system on a particle of mass $m_0$  is F. Then

$\left|F_1\right|=\left|F_2\right|=\left|F_3\right|$

Angles between${\text{F}}_1\hspace{0.17em}\mathrm{and}\hspace{0.17em}{\mathrm{F}}_2,\hspace{0.17em}{\mathrm{F}}_2\hspace{0.17em}\mathrm{and}\hspace{0.17em}{\mathrm{F}}_3\hspace{0.17em}\mathrm{and}\hspace{0.17em}{\mathrm{F}}_3\hspace{0.17em}\mathrm{and}\hspace{0.17em}{\mathrm{F}}_1,\hspace{0.17em}\mathrm{each}\hspace{0.17em}\mathrm{is}\hspace{0.17em}\mathrm{equal}\hspace{0.17em}\mathrm{to}\hspace{0.17em}{120}^0$ .

If $m_0$ is the centre of the triangle, the three forces due to the 3 masses will cancel out and net force will be zero. 

When  $m_0$ is at the mid-point of a side, forces on $m_0$  due to masses at 2 and 3 cancel out each other.

So, force due to m at 1, on $m_0$   is

${\mathrm{F}}_{\mathrm{net}}=\mathrm{G}\frac{{\mathrm{mm}}_0}{{\left(\sqrt{3}\mathrm{L}/2\right)}^2}=\frac{4{\mathrm{Gmm}}_0}{3{\mathrm{L}}^2}$