If the points A(0, 1, 2),B(2, -1, 3),and C(1, | vedclass

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(B) Let the vertices be $A(0, 1, 2)$,$B(2, -1, 3)$,and $C(1, -3, 1)$.Calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2-x_1

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Isosceles right angled

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(B) Let the vertices be $A(0, 1, 2)$,$B(2, -1, 3)$,and $C(1, -3, 1)$.Calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.$AB = \sqrt{(2-0)^2 + (-1-1)^2 + (3-2)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.$BC = \sqrt{(1-2)^2 + (-3 - (-1))^2 + (1-3)^2} = \sqrt{(-1)^2 + (-2)^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.$AC = \sqrt{(1-0)^2 + (-3-1)^2 + (1-2)^2} = \sqrt{1^2 + (-4)^2 + (-1)^2} = \sqrt{1 + 16 + 1} = \sqrt{18} = 3\sqrt{2}$.Since $AB = BC = 3$,the triangle is isosceles.Check for the right-angled condition: $AB^2 + BC^2 = 3^2 + 3^2 = 9 + 9 = 18$,and $AC^2 = (3\sqrt{2})^2 = 18$.Since $AB^2 + BC^2 = AC^2$,the triangle is a right-angled triangle.Therefore,the triangle is an isosceles right-angled triangle.

If the points $A(0, 1, 2)$,$B(2, -1, 3)$,and $C(1, -3, 1)$ are the vertices of a triangle,then the triangle is

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