G(1,0,1) is the centroid of the triangle ABC. | vedclass

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(D) Let the coordinates of $C$ be $(x, y, z)$.Since $G(1,0,1)$ is the centroid of $	riangle ABC$,we have:$\frac{1+3+x}{3} = 1 \implies 4+x = 3 \impli

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$5 BG^2$

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(D) Let the coordinates of $C$ be $(x, y, z)$.Since $G(1,0,1)$ is the centroid of $	riangle ABC$,we have:$\frac{1+3+x}{3} = 1 \implies 4+x = 3 \implies x = -1$$\frac{-4+1+y}{3} = 0 \implies -3+y = 0 \implies y = 3$$\frac{2+0+z}{3} = 1 \implies 2+z = 3 \implies z = 1$So,$C = (-1, 3, 1)$.Now,calculate $AG^2$:$AG^2 = (1-1)^2 + (0-(-4))^2 + (1-2)^2 = 0^2 + 4^2 + (-1)^2 = 16 + 1 = 17$.Calculate $CG^2$:$CG^2 = (-1-1)^2 + (3-0)^2 + (1-1)^2 = (-2)^2 + 3^2 + 0^2 = 4 + 9 = 13$.Thus,$AG^2 + CG^2 = 17 + 13 = 30$.Calculate $BG^2$:$BG^2 = (3-1)^2 + (1-0)^2 + (0-1)^2 = 2^2 + 1^2 + (-1)^2 = 4 + 1 + 1 = 6$.Comparing the values,$AG^2 + CG^2 = 30 = 5 	imes 6 = 5 BG^2$.

$G(1,0,1)$ is the centroid of the triangle $ABC$. If $A=(1,-4,2)$ and $B=(3,1,0)$,then $AG^2+CG^2=$

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