The centroid of the triangle whose vertices a | vedclass

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(B) The centroid $(G)$ of a triangle with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by the formula: $G = (\frac{x_1

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$(\frac{4}{3}, 0, -\frac{1}{3})$

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(B) The centroid $(G)$ of a triangle with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by the formula: $G = (\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3})$ Given vertices are $P(1, -2, 1)$,$Q(2, 3, -1)$,and $R(1, -1, -1)$. Substituting the values: $x$-coordinate $= \frac{1 + 2 + 1}{3} = \frac{4}{3}$ $y$-coordinate $= \frac{-2 + 3 - 1}{3} = \frac{0}{3} = 0$ $z$-coordinate $= \frac{1 - 1 - 1}{3} = -\frac{1}{3}$ Thus,the centroid is $(\frac{4}{3}, 0, -\frac{1}{3})$.

The centroid of the triangle whose vertices are $P(1, -2, 1)$,$Q(2, 3, -1)$,and $R(1, -1, -1)$ is:

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