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(C) The centroid 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 $\left(\frac{x_1+x_2+x_3}{3}, \frac

Vedclass · Accepted Answer

$72$

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(C) The centroid 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 $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$.Given the vertices $(a, 1, 3)$,$(-2, b, -5)$,and $(4, 7, c)$ and the centroid as $(0, 0, 0)$:$\frac{a-2+4}{3} = 0$ $\Rightarrow a+2 = 0$ $\Rightarrow a = -2$$\frac{1+b+7}{3} = 0$ $\Rightarrow b+8 = 0$ $\Rightarrow b = -8$$\frac{3-5+c}{3} = 0$ $\Rightarrow c-2 = 0$ $\Rightarrow c = 2$Therefore,$a^2 + b^2 + c^2 = (-2)^2 + (-8)^2 + (2)^2 = 4 + 64 + 4 = 72$.

If the centroid of the triangle whose vertices are $(a, 1, 3)$,$(-2, b, -5)$,and $(4, 7, c)$ is the origin,then $a^2 + b^2 + c^2 =$

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