Find the coordinates of the vertex ‘C’ of triangle ABC if its centriod is the origin and the vertices A, B are (1, 1, 1) and (–2, 4,1) respectively.

Given two vertices of triangle A(1, 1, 1), B(–2, 4, 1)
Let the third vertex be C = (x, y, z)
Given centriod G = (0, 0, 0)
Centriod
$\begin{array}{l}\mathrm{G}=\left(\frac{{\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3}{3},\frac{{\mathrm{y}}_1+{\mathrm{y}}_2+{\mathrm{y}}_3}{3},\frac{{\mathrm{z}}_1+{\mathrm{z}}_2+{\mathrm{z}}_3}{3}\right)\\ (0,0,0)=\left(\frac{1-2+\mathrm{x}}{3},\frac{1+4+\mathrm{y}}{3},\frac{1+1+\mathrm{z}}{3}\right)\end{array}$
$\begin{array}{l}\frac{\mathrm{x}-1}{3}=0| \begin{array}{r}\frac{5+\mathrm{y}}{3}=0\\ \mathrm{y}+5=0\\ \mathrm{y}=-5\end{array}| \begin{array}{r}\frac{2+\mathrm{z}}{3}=0\\ \mathrm{z}+2=0\\ \mathrm{z}=-2\end{array}\\ \mathrm{x}-1=0\\ \mathrm{x}=1\end{array}$
C = (1, –5, –2)