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?

(This is not likely to be what you're looking for.)
I think in this problem you can use a simpler solution.
Construct point $H$ outside $\overline{AC}$ such that $AGCH$ forms a Parallelogram. We have

*

*$\overline{AH}=\overline{GC}=2$

*$\overline{AG}=2\sqrt3$

*$\overline{GE}=\overline{EH}\Longrightarrow \overline{GH}=\overline{GB}=2\sqrt2$
Since $\overline{AG}^2=\overline{AH}^2+\overline{GH}^2$, we know that $\angle AHG=90^{\circ}$.
Note that $\triangle AGE=\frac{1}2\triangle AGH=\frac{1}2\cdot\frac{1}2\cdot2\cdot2\sqrt2=\sqrt2$.
Therefore $\triangle ABC=6\triangle AGE=6\sqrt2$.

