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(D) Let the vertices of the tetrahedron be $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,$C(x_3, y_3, z_3)$,and $D(x_4, y_4, z_4)$.Since the coordinates of ea

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$\sqrt{29}$

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(D) Let the vertices of the tetrahedron be $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,$C(x_3, y_3, z_3)$,and $D(x_4, y_4, z_4)$.Since the coordinates of each vertex are in arithmetic progression,we have $y_i = x_i + d_i$ and $z_i = x_i + 2d_i$ for $i = 1, 2, 3, 4$.However,the problem implies a common difference $d$ for the coordinates of each vertex. Thus,$y_i = x_i + d$ and $z_i = x_i + 2d$.The centroid $G$ is given by $\left(\frac{\sum x_i}{4}, \frac{\sum y_i}{4}, \frac{\sum z_i}{4}\right) = (2, 3, k)$.From the $x$-coordinate: $\frac{\sum x_i}{4} = 2 \implies \sum x_i = 8$.From the $y$-coordinate: $\frac{\sum (x_i + d)}{4} = 3 \implies \frac{\sum x_i + 4d}{4} = 3 \implies \frac{8 + 4d}{4} = 3 \implies 2 + d = 3 \implies d = 1$.From the $z$-coordinate: $k = \frac{\sum (x_i + 2d)}{4} = \frac{\sum x_i + 8d}{4} = \frac{8 + 8(1)}{4} = \frac{16}{4} = 4$.Thus,the centroid $G$ is $(2, 3, 4)$.The distance of $G$ from the origin $(0, 0, 0)$ is $\sqrt{2^2 + 3^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29}$.

Let $ABCD$ be a tetrahedron in which the coordinates of each of its vertices are in arithmetic progression. If the centroid $G$ of the tetrahedron is $(2, 3, k)$,then the distance of $G$ from the origin is

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