The triangle formed by the points (0, 7, 10), | vedclass

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(D) Let the vertices of the triangle be $A(0, 7, 10)$,$B(-1, 6, 6)$,and $C(-4, 9, 6)$.Using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2

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Right angled Isosceles

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(D) Let the vertices of the triangle be $A(0, 7, 10)$,$B(-1, 6, 6)$,and $C(-4, 9, 6)$.Using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$:$AB = \sqrt{(-1-0)^2 + (6-7)^2 + (6-10)^2} = \sqrt{1 + 1 + 16} = \sqrt{18} = 3\sqrt{2}$$BC = \sqrt{(-4 - (-1))^2 + (9-6)^2 + (6-6)^2} = \sqrt{(-3)^2 + 3^2 + 0^2} = \sqrt{9 + 9 + 0} = \sqrt{18} = 3\sqrt{2}$$AC = \sqrt{(-4-0)^2 + (9-7)^2 + (6-10)^2} = \sqrt{(-4)^2 + 2^2 + (-4)^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$Now,check for the right-angled triangle condition using Pythagoras theorem:$AB^2 + BC^2 = (3\sqrt{2})^2 + (3\sqrt{2})^2 = 18 + 18 = 36$$AC^2 = 6^2 = 36$Since $AB^2 + BC^2 = AC^2$,the triangle is right-angled.Also,since $AB = BC = 3\sqrt{2}$,the triangle is isosceles.Therefore,the triangle is a right-angled isosceles triangle.

The triangle formed by the points $(0, 7, 10), (-1, 6, 6), (-4, 9, 6)$ is

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