If D(2, 1, 0),E(2, 0, 0),and F(0, 1, 0) are m | vedclass

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(B) Let $A \equiv (x_1, y_1, z_1)$,$B \equiv (x_2, y_2, z_2)$,and $C \equiv (x_3, y_3, z_3)$.Since $F(0, 1, 0)$ is the mid-point of $AB$,we have:$x_1

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$\left(\frac{4}{3}, \frac{2}{3}, 0\right)$

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(B) Let $A \equiv (x_1, y_1, z_1)$,$B \equiv (x_2, y_2, z_2)$,and $C \equiv (x_3, y_3, z_3)$.Since $F(0, 1, 0)$ is the mid-point of $AB$,we have:$x_1 + x_2 = 0, y_1 + y_2 = 2, z_1 + z_2 = 0$  $(1)$Since $D(2, 1, 0)$ is the mid-point of $BC$,we have:$x_2 + x_3 = 4, y_2 + y_3 = 2, z_2 + z_3 = 0$  $(2)$Since $E(2, 0, 0)$ is the mid-point of $AC$,we have:$x_3 + x_1 = 4, y_3 + y_1 = 0, z_3 + z_1 = 0$  $(3)$Adding $(1)$,$(2)$,and $(3)$,we get:$2(x_1 + x_2 + x_3) = 8 \implies x_1 + x_2 + x_3 = 4$$2(y_1 + y_2 + y_3) = 4 \implies y_1 + y_2 + y_3 = 2$$2(z_1 + z_2 + z_3) = 0 \implies z_1 + z_2 + z_3 = 0$The centroid of $	riangle ABC$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}ight)$.Substituting the sums,the centroid is $\left(\frac{4}{3}, \frac{2}{3}, 0ight)$.

If $D(2, 1, 0)$,$E(2, 0, 0)$,and $F(0, 1, 0)$ are mid-points of the sides $BC$,$CA$,and $AB$ of $\triangle ABC$,respectively,then the centroid of $\triangle ABC$ is

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