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(C) Let the vertices be $A(2, -1, 1)$,$B(1, -3, -5)$,and $C(3, -4, -4)$.The side lengths are calculated as follows:$|\vec{AB}| = |(1-2)\hat{i} + (-3 -

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Right-angled

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(C) Let the vertices be $A(2, -1, 1)$,$B(1, -3, -5)$,and $C(3, -4, -4)$.The side lengths are calculated as follows:$|\vec{AB}| = |(1-2)\hat{i} + (-3 - (-1))\hat{j} + (-5-1)\hat{k}| = |-\hat{i} - 2\hat{j} - 6\hat{k}| = \sqrt{(-1)^2 + (-2)^2 + (-6)^2} = \sqrt{1 + 4 + 36} = \sqrt{41}$.$|\vec{BC}| = |(3-1)\hat{i} + (-4 - (-3))\hat{j} + (-4 - (-5))\hat{k}| = |2\hat{i} - \hat{j} + \hat{k}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$.$|\vec{CA}| = |(2-3)\hat{i} + (-1 - (-4))\hat{j} + (1 - (-4))\hat{k}| = |-\hat{i} + 3\hat{j} + 5\hat{k}| = \sqrt{(-1)^2 + 3^2 + 5^2} = \sqrt{1 + 9 + 25} = \sqrt{35}$.Since $|\vec{AB}|^2 = 41$ and $|\vec{BC}|^2 + |\vec{CA}|^2 = 6 + 35 = 41$,we have $|\vec{AB}|^2 = |\vec{BC}|^2 + |\vec{CA}|^2$.Thus,the triangle is a right-angled triangle.

If the position vectors of the vertices of a triangle are $2\hat{i} - \hat{j} + \hat{k}$,$\hat{i} - 3\hat{j} - 5\hat{k}$,and $3\hat{i} - 4\hat{j} - 4\hat{k}$,then the triangle is:

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