Let ABCD be a tetrahedron in which the coordi | vedclass

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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 $\sqrt{2^2 + 3^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29}$.

Column $I$	Column $II$
$(A)$ The centroid of the triangle formed by $(2, 3, -1)$,$(5, 6, 3)$,$(2, -3, 1)$ is	$(p)$ $(2, 2, 2)$
$(B)$ The circumcentre of the triangle formed by $(1, 2, 3)$,$(2, 3, 1)$,$(3, 1, 2)$ is	$(q)$ $(3, 1, 4)$
$(C)$ The orthocentre of the triangle formed by $(2, 1, 5)$,$(3, 2, 3)$,$(4, 0, 4)$ is	$(r)$ $(1, 1, 0)$
$(D)$ The incentre of the triangle formed by $(0, 0, 0)$,$(3, 0, 0)$,$(0, 4, 0)$ is	$(s)$ $(3, 2, 1)$

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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