Coordinates of verticies of tetrahedron are in
AP.
Let $A\left(x_1, y_1, z_1\right), B\left(x_2, y_2, z_2\right), C\left(x_3, y_3, z_3\right)$ and $D\left(x_4, y_4, z_4\right)$
Now, each coordinates are in AP
$\begin{aligned} \therefore \quad y_1 & =x_1+d, z_1=x_1+2 d \\ y_2 & =x_2+d, z_2=x_2+2 d \\ y_3 & =x_3+d, z_3=x_3+2 d\end{aligned}$
$\therefore$ centroid of tetrahedron
$\begin{gathered}=\frac{x_1+x_2+x_3+x_4}{4}, \frac{y_1+y_2+y_3+y_4}{4} . \\ \begin{array}{c}z_1+z_2+z_3+z_4 \\ 4\end{array}\end{gathered}$
$\begin{aligned} \therefore \quad 2 & =\frac{x_1+x_2+x_3+x_4}{4} \\ 3 & =\frac{x_1+x_2+x_3+x_1+4 d}{4}\end{aligned}$
and $k=\frac{x_1+x_2+x_3+x_4+8 d}{4}$
$\begin{array}{cc}\therefore & x_1+x_2+x_3+x_4=8, d=1 \\ \Rightarrow & k=\frac{8+8}{4}=4\end{array}$
$\therefore$ Distance of $G$ from the origion
$=\sqrt{2^2+3^2+4^2}=\sqrt{29}$