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(D) Let the coordinates of the vertices of the tetrahedron be $A(a_1-d, a_1, a_1+d)$,$B(a_2-d, a_2, a_2+d)$,$C(a_3-d, a_3, a_3+d)$,and $D(a_4-d, a_4,

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$\sqrt{29}$

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(D) Let the coordinates of the vertices of the tetrahedron be $A(a_1-d, a_1, a_1+d)$,$B(a_2-d, a_2, a_2+d)$,$C(a_3-d, a_3, a_3+d)$,and $D(a_4-d, a_4, a_4+d)$.The centroid $G$ is given by the average of the coordinates of the vertices:$G = \left(\frac{\sum a_i - 4d}{4}, \frac{\sum a_i}{4}, \frac{\sum a_i + 4d}{4}\right) = (2, 3, k)$.Equating the coordinates,we get:$1) \frac{\sum a_i - 4d}{4} = 2 \implies \sum a_i - 4d = 8$$2) \frac{\sum a_i}{4} = 3 \implies \sum a_i = 12$$3) \frac{\sum a_i + 4d}{4} = k \implies \sum a_i + 4d = 4k$Substituting $\sum a_i = 12$ into the first equation: $12 - 4d = 8 \implies 4d = 4 \implies d = 1$.Substituting $\sum a_i = 12$ and $d = 1$ into the third equation: $12 + 4(1) = 4k \implies 16 = 4k \implies k = 4$.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 with the same common difference. If the centroid $G$ of the tetrahedron is $(2, 3, k)$,then the distance of $G$ from the origin is

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