If the vertices of quadrilateral are \((1,1,1),(-2,4,1)\) \((-1,5,5),(2,2,5)\) then it is (A) rectangle (B) square (C) parallelogram (D) rhombus

We will use the distance formula to find the distances between each pair of adjacent vertices of the given quadrilateral. The distance formula is given by: \(D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\) Applying the distance formula for each pair of adjacent vertices, we get: Distance between A(1,1,1) and B(-2,4,1): \(AB = \sqrt{(-2-1)^2 + (4-1)^2 + (1-1)^2} = \sqrt{9 + 9 + 0} = \sqrt{18}\) Distance between B(-2,4,1) and C(-1,5,5): \(BC = \sqrt{(-1+2)^2 + (5-4)^2 + (5-1)^2} = \sqrt{1 + 1 + 16} = \sqrt{18}\) Distance between C(-1,5,5) and D(2,2,5): \(CD = \sqrt{(2+1)^2 + (2-5)^2 + (5-5)^2} = \sqrt{9 + 9 + 0} = \sqrt{18}\) Distance between D(2,2,5) and A(1,1,1): \(DA = \sqrt{(1-2)^2 + (1-2)^2 + (1-5)^2} = \sqrt{1 + 1 + 16} = \sqrt{18}\)

Now we will calculate the distance between the two pairs of opposite vertices of the quadrilateral, which represents the diagonals. We can use the same distance formula as in step 1. Distance between A(1,1,1) and C(-1,5,5): \(AC = \sqrt{(-1-1)^2 + (5-1)^2 + (5-1)^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6\) Distance between B(-2,4,1) and D(2,2,5): \(BD = \sqrt{(2+2)^2 + (2-4)^2 + (5-1)^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6\)

Now that we have the lengths of all sides and diagonals, let's analyze the quadrilateral based on these values: - All four sides have equal lengths (\(AB = BC = CD = DA = \sqrt{18}\)). - Both the diagonals have equal lengths (\(AC = BD = 6\)). Since the quadrilateral has all sides equal in length, it can be either a square, rhombus, or a parallelogram. However, since both diagonals are also equal, it follows the properties of a rectangle. Hence, the given quadrilateral is a (A) rectangle.