Seja BH a altura do triângulo.
[tex3]25^2 = BH^2+7^2 \implies BH = 24\\
HG = \frac{BH}{3} = \frac{24}{3} = 8 \implies BG = 16\\
S_{ABC} = \frac{14.24}{2} = 168\\
p=\frac{14+25+25}{2} = 32\\
r= \frac{S}{p}=\frac{168}{32} = \frac{21}{4}\\
GI = GH-r = 8-\frac{21}{4} = \frac{11}{4}\\
PotG(I) = GI^2 - r^2 = (\frac{11}{4})^2 - (\frac{21}{4} )^2 = \frac{121}{16 }-\frac{441}{16} = -\frac{320}{16} = \boxed{-20}[/tex3]...