As shown above, let the three medians intersect at $G$ which is $\triangle{ABC}$'s centroid. It is easy to show $$S_{\triangle{AFG}}=\dfrac{1}{6}\cdot S_{\triangle{ABC}}$$ Let $M$ be the middle point of $AG$, then we have $$S_{\triangle{FMG}}=\dfrac{1}{2}\cdot S_{\triangle{AFG}}=\dfrac{1}{12}\cdot S_{\triangle{ABC}}$$ Now, consider $\triangle{FMG}$. Note that the lengths of its three sides equal $\dfrac{1}{3}\cdot m_a, \dfrac{1}{3}\cdot m_b,$ and $\dfrac{1}{3}\cdot m_c$, respectively. Let $m=\dfrac{1}{2}(m_a + m_b +m_c)$. By Heron's formula, we have $$S_{\triangle{FMG}}=\sqrt{\dfrac{1}{3}m\cdot\dfrac{1}{3}(m-m_a)\cdot\dfrac{1}{3}(m-m_b)\cdot\dfrac{1}{3}(m-m_c))}=\dfrac{1}{9}\sqrt{m(m-m_a)(m-m_b)(m-m_c)}$$ Hence, we conclude $$S_{\triangle{ABC}}=12\cdot S_{\triangle{FMG}} =\boxed{\dfrac{4}{3}\sqrt{m(m-m_a)(m-m_b)(m-m_c)}}$$