If the centroid of the tetrahedron OABC,where | vedclass

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(A) The centroid of a tetrahedron with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,$(x_3, y_3, z_3)$,and $(x_4, y_4, z_4)$ is given by $\left( \frac{

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$\sqrt{107}$

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(A) The centroid of a tetrahedron with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,$(x_3, y_3, z_3)$,and $(x_4, y_4, z_4)$ is given by $\left( \frac{x_1+x_2+x_3+x_4}{4}, \frac{y_1+y_2+y_3+y_4}{4}, \frac{z_1+z_2+z_3+z_4}{4} \right)$.Given vertices are $O(0, 0, 0)$,$A(a, 2, 3)$,$B(1, b, 2)$,and $C(2, 1, c)$.The centroid is $(1, 2, -1)$.Equating the coordinates:$\frac{a+1+2+0}{4} = 1 \Rightarrow a+3 = 4 \Rightarrow a = 1$.$\frac{2+b+1+0}{4} = 2 \Rightarrow b+3 = 8 \Rightarrow b = 5$.$\frac{3+2+c+0}{4} = -1 \Rightarrow c+5 = -4 \Rightarrow c = -9$.Thus,point $P$ is $(1, 5, -9)$.The distance of $P$ from the origin $(0, 0, 0)$ is $\sqrt{a^2 + b^2 + c^2} = \sqrt{1^2 + 5^2 + (-9)^2} = \sqrt{1 + 25 + 81} = \sqrt{107}$.

If the centroid of the tetrahedron $OABC$,where $O$ is the origin $(0, 0, 0)$ and $A, B, C$ are given by $(a, 2, 3)$,$(1, b, 2)$,and $(2, 1, c)$ respectively,is $(1, 2, -1)$,then the distance of point $P(a, b, c)$ from the origin is equal to:

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