If the vertices of a triangle are (1, 2, 3),( | vedclass

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(D) Let the vertices be $A(1, 2, 3)$,$B(2, 3, 1)$,and $C(3, 1, 2)$.Calculate the side lengths:$AB = \sqrt{(2-1)^2 + (3-2)^2 + (1-3)^2} = \sqrt{1^2 + 1

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$(8, 8, 8)$

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(D) Let the vertices be $A(1, 2, 3)$,$B(2, 3, 1)$,and $C(3, 1, 2)$.Calculate the side lengths:$AB = \sqrt{(2-1)^2 + (3-2)^2 + (1-3)^2} = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{1+1+4} = \sqrt{6}$.$BC = \sqrt{(3-2)^2 + (1-3)^2 + (2-1)^2} = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1+4+1} = \sqrt{6}$.$AC = \sqrt{(3-1)^2 + (1-2)^2 + (2-3)^2} = \sqrt{2^2 + (-1)^2 + (-1)^2} = \sqrt{4+1+1} = \sqrt{6}$.Since $AB = BC = AC = \sqrt{6}$,the triangle is equilateral.For an equilateral triangle,the orthocenter $(H)$,centroid $(G)$,circumcenter $(S)$,and incenter $(I)$ coincide.Thus,$H = G = S = I = \left(\frac{1+2+3}{3}, \frac{2+3+1}{3}, \frac{3+1+2}{3}\right) = (2, 2, 2)$.Therefore,$H+G+S+I = (2, 2, 2) + (2, 2, 2) + (2, 2, 2) + (2, 2, 2) = (8, 8, 8)$.

If the vertices of a triangle are $(1, 2, 3)$,$(2, 3, 1)$,and $(3, 1, 2)$,and if $H, G, S$,and $I$ respectively denote its orthocenter,centroid,circumcenter,and incenter,then $H+G+S+I$ is equal to:

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