\(\left\{{}\begin{matrix}x_G=\dfrac{x_A+x_B+x_C}{3}=\dfrac{1+2+1}{3}=\dfrac{4}{3}\\y_G=\dfrac{y_A+y_B+y_C}{3}=\dfrac{-2-1+1}{3}=-\dfrac{2}{3}\\z_G=\dfrac{z_A+z_B+z_C}{3}=\dfrac{0+1+0}{3}=\dfrac{1}{3}\end{matrix}\right.\Rightarrow G\left(\dfrac{4}{3};-\dfrac{2}{3};\dfrac{1}{3}\right)\)
\(\overrightarrow{CD}\left(-1;-3;0\right)\) la vecto phap tuyen cua mp do
\(\Rightarrow\left(P\right):-1\left(x-\dfrac{4}{3}\right)-3\left(y+\dfrac{2}{3}\right)+0=0\Leftrightarrow x+3y+\dfrac{2}{3}=0\)