<s>[INST] 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?
 [/INST] I think, it's better to use $$S_{\Delta ABC}=\sqrt{p(p-a)(p-b)(p-c)}=$$
$$=\frac{1}{4}\sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}=\frac{1}{4}\sqrt{\sum_{cyc}(2a^2b^2-a^4)}$$ because from the given easy to get $a^2$, $b^2$ and $c^2$.
Indeed, $$\frac{1}{3}\sqrt{2b^2+2c^2-a^2}=2\sqrt3,$$
$$\frac{1}{3}\sqrt{2a^2+2c^2-b^2}=2\sqrt2$$ and
$$\frac{1}{3}\sqrt{2a^2+2b^2-c^2}=2,$$
which gives $$\frac{1}{3}(a^2+b^2+c^2)=4(3+2+1)$$ or
$$a^2+b^2+c^2=72,$$ which gives
$$2(72-a^2)-a^2=108$$ or $$a^2=12.$$
By the similar way we obtain: $b^2=24$ and $c^2=36$, which gives $$S_{\Delta ABC}=6\sqrt2.$$
</s>