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(D) Let the vertices of the triangle be $A(2, 4, -1),$ $B(4, 5, 1),$ and $C(3, 6, -3).$ The side vectors are: $\vec{AB} = (4-2)i + (5-4)j + (1-(-1))k

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Right angled isosceles

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(D) Let the vertices of the triangle be $A(2, 4, -1),$ $B(4, 5, 1),$ and $C(3, 6, -3).$ The side vectors are: $\vec{AB} = (4-2)i + (5-4)j + (1-(-1))k = 2i + j + 2k$ $\vec{BC} = (3-4)i + (6-5)j + (-3-1)k = -i + j - 4k$ $\vec{CA} = (2-3)i + (4-6)j + (-1-(-3))k = -i - 2j + 2k$ Calculating the magnitudes of the sides: $|\vec{AB}| = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$ $|\vec{BC}| = \sqrt{(-1)^2 + 1^2 + (-4)^2} = \sqrt{1 + 1 + 16} = \sqrt{18} = 3\sqrt{2}$ $|\vec{CA}| = \sqrt{(-1)^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$ Since $|\vec{AB}| = |\vec{CA}| = 3,$ the triangle is isosceles. Checking for the right-angled condition: $|\vec{AB}|^2 + |\vec{CA}|^2 = 3^2 + 3^2 = 9 + 9 = 18$ $|\vec{BC}|^2 = (\sqrt{18})^2 = 18$ Since $|\vec{AB}|^2 + |\vec{CA}|^2 = |\vec{BC}|^2,$ the triangle satisfies the Pythagorean theorem. Therefore,the triangle is a right-angled isosceles triangle.

If the position vectors of the vertices of a triangle are $2i + 4j - k,$ $4i + 5j + k,$ and $3i + 6j - 3k,$ then the triangle is

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