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(D) 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

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$34$

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(D) 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 $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$.For the given vertices $(4, p, -3)$,$(-1, -1, 2)$,and $(3, 5, -8)$,the centroid is $\left(\frac{4-1+3}{3}, \frac{p-1+5}{3}, \frac{-3+2-8}{3}\right) = \left(2, \frac{p+4}{3}, -3\right)$.The mid-point of $(1, 4, -2)$ and $(q, 2, -4)$ is $\left(\frac{1+q}{2}, \frac{4+2}{2}, \frac{-2-4}{2}\right) = \left(\frac{1+q}{2}, 3, -3\right)$.Equating the coordinates of the centroid and the mid-point:$2 = \frac{1+q}{2} \Rightarrow 4 = 1+q \Rightarrow q = 3$.$\frac{p+4}{3} = 3 \Rightarrow p+4 = 9 \Rightarrow p = 5$.Thus,$p^2 + q^2 = 5^2 + 3^2 = 25 + 9 = 34$.

If the centroid of a triangle with vertices $(4, p, -3)$,$(-1, -1, 2)$,and $(3, 5, -8)$ is given by the mid-point of $(1, 4, -2)$ and $(q, 2, -4)$,then $p^2 + q^2 =$

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