<s>[INST] Find the area of ​the shaded region in the triangle below? For reference: In figure $G$ is the centroid of the triangle $ABC$; if the area of ​​the $FGC$ triangle is $9m^2$, the area of ​​the FGB triangle is $16m^2$
Calculate the area of ​​the shaded region. (Answer:$7m^2$)
If possible by geometry
My progress:

$S_{FGC} = \frac{b.h_1}{2} = \frac{FG.h_1}{2}\implies FG = \frac{18}{h_1}\\
S_{FGB}=\frac{b.h_2}{2} = \frac{FG.h_2}{2} \implies FG = \frac{32}{h_2}\\
\therefore \frac{18}{h_1} = \frac{32}{h_2}\implies \frac{h_1}{h_2} = \frac{32}{18}=\frac{16}{9}\\
S_{ABG} = S_{BCG} = S_{ACG}$
...??? I'm not able to develop this
 [/INST] With the usual conventions:
\begin{aligned}
\overrightarrow{GF}\times\overrightarrow{GB} &= (0,0,32)\\
\overrightarrow{GF}\times\overrightarrow{GC} &= (0,0,-18)\\
\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC} &= \overrightarrow{0}
\end{aligned}
The area you search:
\begin{multline*}
\frac{1}{2}\big\lvert\overrightarrow{GF}\times\overrightarrow{GA}\big\rvert=\frac{1}{2}\Big\lvert\overrightarrow{GF}\times\big(-\overrightarrow{GB}-\overrightarrow{GC}\big)\Big\rvert
=\frac{1}{2}\big\lvert-\overrightarrow{GF}\times\overrightarrow{GB}-\overrightarrow{GF}\times\overrightarrow{GC}\big\rvert=\frac{1}{2}\big\lvert(0,0,-14)\big\rvert = 7
\end{multline*}
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