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

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(B) Let the vertices of $	riangle ABC$ be $A(x_1, y_1, z_1)$, $B(x_2, y_2, z_2)$, and $C(x_3, y_3, z_3)$.Since $D(2, 1, 0)$ is the mid-point of $BC$,

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

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(B) Let the vertices of $	riangle ABC$ be $A(x_1, y_1, z_1)$, $B(x_2, y_2, z_2)$, and $C(x_3, y_3, z_3)$.Since $D(2, 1, 0)$ is the mid-point of $BC$, we have:$\frac{x_2+x_3}{2} = 2 \implies x_2+x_3 = 4$$\frac{y_2+y_3}{2} = 1 \implies y_2+y_3 = 2$$\frac{z_2+z_3}{2} = 0 \implies z_2+z_3 = 0$Since $E(2, 0, 0)$ is the mid-point of $AC$, we have:$\frac{x_1+x_3}{2} = 2 \implies x_1+x_3 = 4$$\frac{y_1+y_3}{2} = 0 \implies y_1+y_3 = 0$$\frac{z_1+z_3}{2} = 0 \implies z_1+z_3 = 0$Since $F(0, 1, 0)$ is the mid-point of $AB$, we have:$\frac{x_1+x_2}{2} = 0 \implies x_1+x_2 = 0$$\frac{y_1+y_2}{2} = 1 \implies y_1+y_2 = 2$$\frac{z_1+z_2}{2} = 0 \implies z_1+z_2 = 0$Adding these equations:$2(x_1+x_2+x_3) = 4+4+0 = 8 \implies x_1+x_2+x_3 = 4$$2(y_1+y_2+y_3) = 2+0+2 = 4 \implies y_1+y_2+y_3 = 2$$2(z_1+z_2+z_3) = 0+0+0 = 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 the mid-points of the sides $BC$, $CA$, and $AB$ of $\triangle ABC$, respectively, then the centroid of $\triangle ABC$ is:

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