Using the first 2 equations gives $$2(xy+yz+zx) = 24^2-210=366$$So $xy+yz+zx=183$. We then see by Vieta that $a,b,c$ are solutions to the cubic $$x^3-24x^2+183x-440=0$$Giving that $(a,b,c)=(5,8,11)$ and its cyclic permutations.

Let $ a, b, c >0 , a + b + c =24$ and $ a^2 + b^2 + c^2 =210 .$ Prove that $$440-6\sqrt{3} \leq abc \leq 440+6\sqrt{3}$$