If G(3, -5, r) is the centroid of triangle AB | vedclass

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(D) 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 $(\frac{x_1+x_2+x_3}{3}, \frac{

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$-2, 3, 2$

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(D) 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 $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3})$.Given $A(7, -8, 1)$,$B(p, q, 5)$,$C(q+1, 5p, 0)$,and $G(3, -5, r)$.Equating the coordinates:$x$-coordinate: $\frac{7+p+q+1}{3} = 3 \Rightarrow p+q+8 = 9 \Rightarrow p+q = 1$ (Equation $1$)$y$-coordinate: $\frac{-8+q+5p}{3} = -5 \Rightarrow 5p+q-8 = -15 \Rightarrow 5p+q = -7$ (Equation $2$)$z$-coordinate: $\frac{1+5+0}{3} = r \Rightarrow r = \frac{6}{3} = 2$Subtracting Equation $1$ from Equation $2$: $(5p+q) - (p+q) = -7 - 1 \Rightarrow 4p = -8 \Rightarrow p = -2$.Substituting $p = -2$ into Equation $1$: $-2 + q = 1 \Rightarrow q = 3$.Thus,the values are $p = -2, q = 3, r = 2$.

If $G(3, -5, r)$ is the centroid of triangle $ABC$ where $A(7, -8, 1)$,$B(p, q, 5)$,and $C(q+1, 5p, 0)$ are the vertices of the triangle,then the values of $p, q, r$ are respectively . . . . . . .

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