Vertices of $\triangle P Q R$ are $P(2 a, 2,6)$, $Q(-4,3 b,-10)$ and $R(8,14,2 c)$.
Centroid of the $\triangle P Q R$ is
$\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$
or $\left(\frac{2 a-4+8}{3}, \frac{2+3 b+14}{3}, \frac{6-10+2 c}{3}\right)$
or $\left(\frac{2 a+4}{3}, \frac{3 b+16}{3}, \frac{2 c-4}{3}\right)$
We are given that centroid is origin i.e $(0,0,0)$
$\begin{array}{ll}
\Rightarrow \frac{2 a+4}{3}=0 & \therefore a=-2 \\
\Rightarrow \frac{3 b+16}{3}=0 & \therefore b=\frac{-16}{3} \\
\Rightarrow \frac{2 c-4}{3}=0 & \therefore c=2
\end{array}$
Thus, $a=-2, b=-\frac{16}{3}, c=2$