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(B) Let the vertices be $A(1, 1, 1)$,$B(-2, 4, 1)$,$C(-1, 5, 5)$,and $D(2, 2, 5)$.Calculate the lengths of the sides using the distance formula $d = \

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Square

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(B) Let the vertices be $A(1, 1, 1)$,$B(-2, 4, 1)$,$C(-1, 5, 5)$,and $D(2, 2, 5)$.Calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$:$AB = \sqrt{(-2-1)^2 + (4-1)^2 + (1-1)^2} = \sqrt{(-3)^2 + 3^2 + 0^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$.$BC = \sqrt{(-1-(-2))^2 + (5-4)^2 + (5-1)^2} = \sqrt{1^2 + 1^2 + 4^2} = \sqrt{1+1+16} = \sqrt{18} = 3\sqrt{2}$.$CD = \sqrt{(2-(-1))^2 + (2-5)^2 + (5-5)^2} = \sqrt{3^2 + (-3)^2 + 0^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$.$DA = \sqrt{(1-2)^2 + (1-2)^2 + (1-5)^2} = \sqrt{(-1)^2 + (-1)^2 + (-4)^2} = \sqrt{1+1+16} = \sqrt{18} = 3\sqrt{2}$.Since all sides are equal,we check the diagonals $AC$ and $BD$:$AC = \sqrt{(-1-1)^2 + (5-1)^2 + (5-1)^2} = \sqrt{(-2)^2 + 4^2 + 4^2} = \sqrt{4+16+16} = \sqrt{36} = 6$.$BD = \sqrt{(2-(-2))^2 + (2-4)^2 + (5-1)^2} = \sqrt{4^2 + (-2)^2 + 4^2} = \sqrt{16+4+16} = \sqrt{36} = 6$.Since all sides are equal $(3\sqrt{2})$ and both diagonals are equal $(6)$,the figure is a square.

Points $(1, 1, 1), (-2, 4, 1), (-1, 5, 5)$ and $(2, 2, 5)$ are the vertices of a

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