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(C) Let the position vectors of the vertices be $\vec{A} = 6i + 4j + 5k$,$\vec{B} = 4i + 5j + 6k$,and $\vec{C} = 5i + 6j + 4k$. To find the lengths of

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Equilateral

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(C) Let the position vectors of the vertices be $\vec{A} = 6i + 4j + 5k$,$\vec{B} = 4i + 5j + 6k$,and $\vec{C} = 5i + 6j + 4k$. To find the lengths of the sides,we calculate the magnitudes of the vectors $\vec{AB}$,$\vec{BC}$,and $\vec{CA}$. $\vec{AB} = \vec{B} - \vec{A} = (4-6)i + (5-4)j + (6-5)k = -2i + j + k$. The length $AB = |\vec{AB}| = \sqrt{(-2)^2 + 1^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$. $\vec{BC} = \vec{C} - \vec{B} = (5-4)i + (6-5)j + (4-6)k = i + j - 2k$. The length $BC = |\vec{BC}| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$. $\vec{CA} = \vec{A} - \vec{C} = (6-5)i + (4-6)j + (5-4)k = i - 2j + k$. The length $CA = |\vec{CA}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$. Since $AB = BC = CA = \sqrt{6}$,the triangle is an equilateral triangle.

If the position vectors of the vertices of a triangle are $6i + 4j + 5k$,$4i + 5j + 6k$,and $5i + 6j + 4k$,then the triangle is

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