Neptune dance

As, the area of the triangle formed by the planets, needs to be constant at every moment under the influence of mutual gravitational force, it can be safely concluded that, the mass of each planet needs to be same i.e. $m=\frac{3*10^{26}}{3}=1*10^{26}kg.$ and they need to keep themselves at equal distances; for stability under symmetrical action of mutual gravitational forces, thus forming an equilateral triangle with its centroid coinciding with the center-of-mass of the system along which the axis passes, having an area $(A)$ of $1\text{ A.U.}^2=(149.6*10^9)^2\,m^2$ . Let's say, the length of each side = distance between the planets= $"a"$ , and the circum-radius of the triangle=radius of the orbit= $"r"$ . So, it follows that $r=\frac{a}{\sqrt{3}}$ ,and $A=\frac{\sqrt{3}*(a^2)}{4}$ . The forces acting on each of the planets, are (i) Centripetal force (i.e. $mr\omega^2$ ) and (ii) The gravitational forces (i.e $\frac{G*m*m}{a^2}$ due to the other two planets acting along the adjacent sides of the triangle with the planet at their common end, which are at an angular separation of $-30^{\circ}$ & $30^{\circ}$ from the line of action of centripetal force. Therefore, taking the components of forces along the line of action of centripetal force, and considering the fact that, the forces are in equilibrium, it can be written & simplified as follows: $mr\omega^2=\frac{G*m*m}{a^2}*\cos{30^{\circ}}+\frac{G*m*m}{a^2}*\cos{(-30)^{\circ}}$ $\Rightarrow\omega=\sqrt{\frac{3*G*m}{a^3}}=\sqrt{\frac{3*G*m}{{(\frac{4*A}{\sqrt{3}})}^\frac{3}{2}}}.$ On substituting the values, we get $\text{Angular Velocity}=\omega=1.3*10^{-9} rad/s.$ $\Rightarrow\omega\approx1*10^{-9} rad/s.$ (after rounding up as mentioned in the problem.) N.B - All the notations denote their usual meanings, unless otherwise specified

I've devised my own system based on the information given:-

-The mass of all three planets are equal

-The planets are placed at the three corners of an equilateral triangle

Thus, the side and altitude of the triangle can be calculated from Area as:-

-side(S)=sqrt{4*(Area) /sqrt{3}}

-Altitude(A)=(side)/sqrt{3}

-The radius(r) of the orbit of any one of the planets is equal to 2/3 times the altitude of the triangle.

-This implies, r = 2(A)/3

-let the angular velocity of planet be 'w' and mass be 'm'.

-We can say that the net gravitational attraction of the two planets provides the necessary centripetal acceleration to the third planet.

-Therefore, (w^2)r=Gm/(S^2)

-Thus w = sqrt{Gm/r(S^2)}