Given that $G$ is the centroid of $\Delta ABC$, $GA = 2\sqrt{3}$ , $GB = 2\sqrt{2}$, $GC = 2$ . Find $[\Delta ABC]$. 
Given that $G$ is the centroid of $\Delta ABC$, $GA = 2\sqrt{3}$ , $GB = 2\sqrt{2}$, $GC = 2$ . Find $[\Delta ABC]$.

What I Tried: Here is a picture :-

I know the centroid divides each of the medians in the ratio $2:1$ . So $AD = 3\sqrt{3}$ , $BE = 3\sqrt{2}$ , $CF = 3$ . 
From this site :- https://mathworld.wolfram.com/TriangleMedian.html, I find that the area of the triangle will be :- $$\frac{4}{3}\sqrt{s_m(s_m - m_1)(s_m - m_2)(s_m - m_3)}$$
Where $m_1,m_2,m_3$ are the medians of the triangle and $s_m = \frac{m_1 + m_2 + m_3}{2}$ .
After putting the respective values for the medians I get that $[\Delta ABC]$ is :-
$$\frac{4}{3}\sqrt{\Bigg(\frac{3(\sqrt{3} + \sqrt{2} + 1)}{2}\Bigg)\Bigg(\frac{3(\sqrt{2} + 1 - \sqrt{3})}{2}\Bigg)\Bigg(\frac{3(\sqrt{3} + 1 - \sqrt{2})}{2}\Bigg)\Bigg(\frac{3(\sqrt{3} + \sqrt{2} - 1)}{2}\Bigg)}$$
$$\rightarrow \frac{4}{3}\sqrt{\frac{81(\sqrt{3} + \sqrt{2} + 1)(\sqrt{2} + 1 - \sqrt{3})(\sqrt{3} + 1 - \sqrt{2})(\sqrt{3} + \sqrt{2} - 1)}{16}}$$
I am almost to the answer (assuming I made no mistake), but I think this simplification is getting complicated. How do I proceed next?
Can anyone help me?

It is indeed a theorem that can be generalized, although Mathworlddoes n't say so explicitly.
The area of a triangle formed by medians ( computed for example from Brahmagupta/Heron formula ) is three-fourths the area formed by the corresponding sides of the given triangle.
It can also be proved by projective geometry.
Linear scale $k=\sin \frac{\pi}{3}$ can be established from relative proportions of the simplest equilateral triangle.
In our case  hypothetical medians after scaling up full sides from centroid by $\text{50%} :  3(\sqrt 3, \sqrt 2,1)$ calculates to $\dfrac{9}{\sqrt 2};$
So the circumscribed triangle area would be:
$$ \dfrac{{\dfrac{9}{\sqrt 2}}} {\sin^2\dfrac{\pi}{3}} = 6 \sqrt 2.$$
