Given, mid points of the sides of triangle are $(1,5,-1),(0,4,-2)$ and $(2,3,4)$.
Let, $\left(x_{1}, y_{1}, z_{1}\right)=(1,5,=1)$,
$\left(x_{2}, y_{2}, z_{2}\right)=(0,4,-2) \text {, }$
$\left(x_{3}, y_{3}, z_{3}\right)=(2,3,4) \text {. }$
Then centroid of triangle, $G(x, y, z)=$
$\begin{aligned}
&\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}, \frac{z_{1}+z_{2}+z_{3}}{3}\right) \\
&G(x, y, z)=\left(\frac{1+0+2}{3}, \frac{5+4+3}{3}, \frac{-1-2+4}{3}\right) \\
&=\left(1,4, \frac{1}{3}\right)
\end{aligned}$