Let $\quad A \equiv\left(x_1, y_1, z_1\right), \quad B \equiv\left(x_2, y_2, z_2\right) \quad$ and $C \equiv\left(x_3, y_3, z_3\right)$


Since, $F$ is the mid-point of $A B$.
$$
\left.\therefore \quad \begin{array}{l}
x_1+x_2=0 \\
y_1+y_2=2 \\
z_1+z_2=0
\end{array}\right\}
$$
Since, $D$ is the mid-point of $B C$.
$$
\left.\therefore \quad \begin{array}{l}
x_2+x_3=4 \\
y_2+y_3=2 \\
z_2+z_3=0
\end{array}\right\}
$$
and $E$ is the mid-point of $A C$
$$
\left.\therefore \quad \begin{array}{l}
x_3+x_1=4 \\
y_3+y_1=0 \\
z_3+z_1=0
\end{array}\right\}
$$
So,
$$
\left.\begin{array}{l}
x_1+x_2+x_3=4 \\
y_1+y_2+y_3=2 \\
z_1+z_2+z_3=0
\end{array}\right\}
$$
$$
\begin{array}{llll}
\therefore & x_3=4, & y_3=0, & z_3=0 \\
& x_1=0, & y_1=0, & z_1=0 \\
\text { and } & x_2=0, & y_2=2, & z_2=0
\end{array}
$$
$\therefore$ Centroid of $\triangle A B C$
$$
\begin{aligned}
& =\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right) \\
& =\left(\frac{4}{3}, \frac{2}{3}, 0\right)
\end{aligned}
$$