12. As shown in the figure, given that the diagonals $A D, B E, C F$ of a convex hexagon intersect at $O$, and the angles between each pair are $60^{\circ}$. If $A B=1, B C=2 \sqrt{2}$, $C D=D E=5, E F=4, F A=3$, then the maximum value of the area $S$ of the convex hexagon is $\qquad$
Given $O A=x, O B=y, O C=z, O D=s, O E=t, O F=u$, then

Thus $S=\frac{\sqrt{3}}{4}(x y+y z+z s+s t+t u+u x)=\frac{\sqrt{3}}{2}(x y+z s+t u)$
$$
\leqslant \frac{\sqrt{3}}{4}\left(x^{2}+y^{2}+z^{2}+s^{2}+t^{2}+u^{2}\right)=\frac{\sqrt{3}}{4}(x y+1+z s+25+t u+16)
$$
$=\frac{\sqrt{3}}{4}(x y+z s+t u)+\frac{21 \sqrt{3}}{2} \Rightarrow \frac{\sqrt{3}}{4}(x y+z s+t u) \leqslant \frac{21 \sqrt{3}}{2} \Rightarrow S \leqslant 21 \sqrt{3}$, equality holds when $x=y=1, z=s=5, t=u=4$. Therefore, the maximum value of the hexagon area $S$ is $21 \sqrt{3}$. Note: This problem is incorrect, under the condition that $x, y, z, s, t, u$ are not all equal, it is impossible to construct the condition $A B^{2}+$ $C D^{2}+E F^{2}=B C^{2}+D E^{2}+F A^{2}$, but the problem-solving approach is worth learning.