Solve the following :

Let us assume that the center of mass of spheres are located at point $P, Q$ and $R$. So, re-drawing the figure, it is an equilateral triangle.

Force on mass $M$ at point $P$

$\mathrm{F}_{\mathrm{PR}}=\mathrm{F}_{\mathrm{PQ}} \frac{G M M}{(2 a)^{2}}=\frac{G M^{2}}{4 a^{2}}=\mathrm{k}($ let $)$

Resultant

$\mathrm{F}_{\mathrm{R}}=\sqrt{F_{P Q}^{2}+F_{P R}^{2}+2 F_{P Q} \cdot F_{P R} \cdot \cos 60^{\circ}}$

$\mathrm{F}_{\mathrm{R}}=\sqrt{k^{2}+k^{2}+2 k^{2} \cdot\left(\frac{1}{2}\right)}$

$\mathrm{F}_{\mathrm{R}}=\sqrt{3} k$

$\mathrm{F}_{\mathrm{R}}=\sqrt{3} \frac{G M^{2}}{4 a^{2}}$