If A(1,4,2) and C(5,-7,1) are two vertices of | vedclass

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(D) Let the coordinates of vertex $B$ be $(x_1, y_1, z_1)$.The centroid $G$ of a triangle with vertices $A(x_A, y_A, z_A)$,$B(x_B, y_B, z_B)$,and $C(x

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$\left(\frac{3}{2},-2,-2\right)$

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(D) Let the coordinates of vertex $B$ be $(x_1, y_1, z_1)$.The centroid $G$ of a triangle with vertices $A(x_A, y_A, z_A)$,$B(x_B, y_B, z_B)$,and $C(x_C, y_C, z_C)$ is given by $\left(\frac{x_A+x_B+x_C}{3}, \frac{y_A+y_B+y_C}{3}, \frac{z_A+z_B+z_C}{3}\right)$.Given $A(1,4,2)$,$C(5,-7,1)$,and $G\left(\frac{4}{3}, 0, \frac{-2}{3}\right)$,we have:$\frac{1+x_1+5}{3} = \frac{4}{3} \Rightarrow 6+x_1 = 4 \Rightarrow x_1 = -2$$\frac{4+y_1-7}{3} = 0 \Rightarrow y_1-3 = 0 \Rightarrow y_1 = 3$$\frac{2+z_1+1}{3} = \frac{-2}{3} \Rightarrow 3+z_1 = -2 \Rightarrow z_1 = -5$So,$B = (-2, 3, -5)$.The midpoint of side $BC$ is $\left(\frac{x_B+x_C}{2}, \frac{y_B+y_C}{2}, \frac{z_B+z_C}{2}\right)$.Midpoint $= \left(\frac{-2+5}{2}, \frac{3-7}{2}, \frac{-5+1}{2}\right) = \left(\frac{3}{2}, -2, -2\right)$.

If $A(1,4,2)$ and $C(5,-7,1)$ are two vertices of triangle $ABC$ and $G\left(\frac{4}{3}, 0, \frac{-2}{3}\right)$ is the centroid of the triangle $ABC$,then the midpoint of side $BC$ is

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