Let two vertices of triangle ABC be (2,4,6) a | vedclass

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(B) Let the vertices of the triangle be $A(2,4,6)$,$B(0,-2,-5)$,and $C(x,y,z)$.Given the centroid $G(2,1,-1)$,we use the formula for the centroid: $G

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$74$

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(B) Let the vertices of the triangle be $A(2,4,6)$,$B(0,-2,-5)$,and $C(x,y,z)$.Given the centroid $G(2,1,-1)$,we use the formula for the centroid: $G = (\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3})$.$\frac{2+0+x}{3} = 2 \Rightarrow 2+x = 6 \Rightarrow x = 4$.$\frac{4-2+y}{3} = 1 \Rightarrow 2+y = 3 \Rightarrow y = 1$.$\frac{6-5+z}{3} = -1 \Rightarrow 1+z = -3 \Rightarrow z = -4$.So,the third vertex is $C(4,1,-4)$.Now,find the image $(\alpha, \beta, \gamma)$ of point $C(4,1,-4)$ in the plane $x+2y+4z-11=0$.The formula for the image of a point $(x_0, y_0, z_0)$ in the plane $ax+by+cz+d=0$ is $\frac{\alpha-x_0}{a} = \frac{\beta-y_0}{b} = \frac{\gamma-z_0}{c} = -2 \frac{ax_0+by_0+cz_0+d}{a^2+b^2+c^2}$.Substituting the values: $\frac{\alpha-4}{1} = \frac{\beta-1}{2} = \frac{\gamma+4}{4} = -2 \frac{4+2(1)+4(-4)-11}{1^2+2^2+4^2} = -2 \frac{4+2-16-11}{1+4+16} = -2 \frac{-21}{21} = 2$.Thus,$\alpha-4 = 2 \Rightarrow \alpha = 6$; $\beta-1 = 4 \Rightarrow \beta = 5$; $\gamma+4 = 8 \Rightarrow \gamma = 4$.The image is $(6,5,4)$.Finally,$\alpha \beta+\beta \gamma+\gamma \alpha = (6)(5) + (5)(4) + (4)(6) = 30 + 20 + 24 = 74$.

Let two vertices of triangle $ABC$ be $(2,4,6)$ and $(0,-2,-5)$,and its centroid be $(2,1,-1)$. If the image of the third vertex in the plane $x+2y+4z=11$ is $(\alpha, \beta, \gamma)$,then $\alpha \beta+\beta \gamma+\gamma \alpha$ is equal to

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