If $\left(3,2,-1\right),\left(4,1,1\right)$ and $\left(6,2,5\right)$are three vertices and $\left(4,2,2\right)$ is the centriod of a tetrahedron, find the fourth vertex.

Given vertices of tetrahedron are
A(3, 2, –1), B(4, 1, 1), C(6, 2, 5)
Let fourth vertex D(x, y, z)
Given G = (4, 2, 2)
Centriod of tetrahedron ‘G’
$\begin{array}{c}G=\left(\frac{{\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3+{\mathrm{x}}_4}{4},\frac{{\mathrm{y}}_1+{\mathrm{y}}_2+{\mathrm{y}}_3+{\mathrm{y}}_4}{4},\frac{z_1+{\mathrm{z}}_2+{\mathrm{z}}_3+{\mathrm{z}}_4}{4}\right)\\ (4,2,2)=\left(\frac{3+4+6+\mathrm{x}}{4},\frac{2+1+2+\mathrm{y}}{4},\frac{-1+1+5+\mathrm{z}}{4}\right)\\ (4,2,2)=\left(\frac{13+\mathrm{x}}{4},\frac{5+\mathrm{y}}{4},\frac{5+\mathrm{z}}{4}\right)\end{array}$

It gives that 

$13+x=16\Rightarrow x=3$

$5+y=8\Rightarrow y=3$

$5+z=8\Rightarrow z=3$

Therefore, fourth vertex is $\boxed{\mathbf{D}\left(3,3,3\right)}$