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(D) Let the vertices of the triangle be $A = (1, 0, 0)$,$B = (0, 1, 0)$,and $C = (0, 0, 1)$.Using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y

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$3 \sqrt{2}$

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(D) Let the vertices of the triangle be $A = (1, 0, 0)$,$B = (0, 1, 0)$,and $C = (0, 0, 1)$.Using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$,we calculate the lengths of the sides:$AB = \sqrt{(0-1)^2 + (1-0)^2 + (0-0)^2} = \sqrt{(-1)^2 + 1^2 + 0^2} = \sqrt{1+1} = \sqrt{2}$.$BC = \sqrt{(0-0)^2 + (0-1)^2 + (1-0)^2} = \sqrt{0^2 + (-1)^2 + 1^2} = \sqrt{1+1} = \sqrt{2}$.$CA = \sqrt{(1-0)^2 + (0-0)^2 + (0-1)^2} = \sqrt{1^2 + 0^2 + (-1)^2} = \sqrt{1+1} = \sqrt{2}$.The perimeter of the triangle is the sum of the lengths of its sides:$	ext{Perimeter} = AB + BC + CA = \sqrt{2} + \sqrt{2} + \sqrt{2} = 3\sqrt{2}$.

The perimeter of the triangle with vertices at $(1,0,0), (0,1,0)$ and $(0,0,1)$ is

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