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(C) The coordinates of the centroid $G$ of a triangle with vertices $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$ are given by $\left(\

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$\left(1, \frac{7}{2}, \frac{3}{2}\right)$

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(C) The coordinates of the centroid $G$ of a triangle with vertices $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$ are given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$.Given $A(4, x, 1)$,$B(y, -5, 2)$,$C(7, 8, 3)$,and $G(3, 5, 2)$,we have:$G = \left(\frac{4+y+7}{3}, \frac{x-5+8}{3}, \frac{1+2+3}{3}\right) = (3, 5, 2)$Equating the coordinates:$\frac{11+y}{3} = 3 \Rightarrow 11+y = 9 \Rightarrow y = -2$$\frac{x+3}{3} = 5 \Rightarrow x+3 = 15 \Rightarrow x = 12$Since $CG$ is a median,$F$ is the midpoint of $AB$. The coordinates of $F$ are:$F = \left(\frac{4+y}{2}, \frac{x-5}{2}, \frac{1+2}{2}\right)$Substituting $x=12$ and $y=-2$:$F = \left(\frac{4-2}{2}, \frac{12-5}{2}, \frac{3}{2}\right) = \left(\frac{2}{2}, \frac{7}{2}, \frac{3}{2}\right) = \left(1, \frac{7}{2}, \frac{3}{2}\right)$.

Let the centroid of a triangle formed by the points $A(4, x, 1)$,$B(y, -5, 2)$ and $C(7, 8, 3)$ be $G(3, 5, 2)$ and $CG$ meet $AB$ in $F$. Then,$F=$

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