The quadrilateral formed by the points A(1,2, | vedclass

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(C) Let the vertices be $A(1,2,5), B(-1,6,1), C(3,4,-3)$ and $D(5,0,1)$.First,we calculate the lengths of the sides using the distance formula $d = \s

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Square

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(C) Let the vertices be $A(1,2,5), B(-1,6,1), C(3,4,-3)$ and $D(5,0,1)$.First,we 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{(-1-1)^2 + (6-2)^2 + (1-5)^2} = \sqrt{(-2)^2 + 4^2 + (-4)^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$$BC = \sqrt{(3-(-1))^2 + (4-6)^2 + (-3-1)^2} = \sqrt{4^2 + (-2)^2 + (-4)^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$$CD = \sqrt{(5-3)^2 + (0-4)^2 + (1-(-3))^2} = \sqrt{2^2 + (-4)^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$$DA = \sqrt{(1-5)^2 + (2-0)^2 + (5-1)^2} = \sqrt{(-4)^2 + 2^2 + 4^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$Since all sides are equal $(AB = BC = CD = DA = 6)$,the quadrilateral is a rhombus.Now,we check the diagonals:$AC = \sqrt{(3-1)^2 + (4-2)^2 + (-3-5)^2} = \sqrt{2^2 + 2^2 + (-8)^2} = \sqrt{4 + 4 + 64} = \sqrt{72} = 6\sqrt{2}$$BD = \sqrt{(5-(-1))^2 + (0-6)^2 + (1-1)^2} = \sqrt{6^2 + (-6)^2 + 0^2} = \sqrt{36 + 36 + 0} = \sqrt{72} = 6\sqrt{2}$Since the diagonals are also equal $(AC = BD = 6\sqrt{2})$,the quadrilateral is a square.

The quadrilateral formed by the points $A(1,2,5), B(-1,6,1), C(3,4,-3)$ and $D(5,0,1)$ is a

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