The points A(2, 3, 5),B(-1, 5, -1),and C(4, - | vedclass

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(D) Let the points be $A(2, 3, 5)$,$B(-1, 5, -1)$,and $C(4, -3, 2)$.Calculate the squared distances between the points using the distance formula $d^2

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an isosceles right angled triangle

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(D) Let the points be $A(2, 3, 5)$,$B(-1, 5, -1)$,and $C(4, -3, 2)$.Calculate the squared distances between the points using the distance formula $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$:$AB^2 = (-1 - 2)^2 + (5 - 3)^2 + (-1 - 5)^2 = (-3)^2 + (2)^2 + (-6)^2 = 9 + 4 + 36 = 49$.$BC^2 = (4 - (-1))^2 + (-3 - 5)^2 + (2 - (-1))^2 = (5)^2 + (-8)^2 + (3)^2 = 25 + 64 + 9 = 98$.$AC^2 = (4 - 2)^2 + (-3 - 3)^2 + (2 - 5)^2 = (2)^2 + (-6)^2 + (-3)^2 = 4 + 36 + 9 = 49$.Since $AB^2 = AC^2 = 49$,the triangle is isosceles because $AB = AC = 7$.Check for the right-angled condition: $AB^2 + AC^2 = 49 + 49 = 98 = BC^2$.Since the sum of the squares of two sides equals the square of the third side,the triangle is a right-angled triangle.Thus,the triangle is an isosceles right-angled triangle.

The points $A(2, 3, 5)$,$B(-1, 5, -1)$,and $C(4, -3, 2)$ form:

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