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(D) Let the vertices of the triangle be $A(2, -1, 1)$,$B(1, -3, -5)$,and $C(3, -4, -4)$.First,we find the vectors representing the sides:$\vec{AB} = (

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right angled

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(D) Let the vertices of the triangle be $A(2, -1, 1)$,$B(1, -3, -5)$,and $C(3, -4, -4)$.First,we find the vectors representing the sides:$\vec{AB} = (1-2)\hat{i} + (-3 - (-1))\hat{j} + (-5-1)\hat{k} = -\hat{i} - 2\hat{j} - 6\hat{k}$$\vec{BC} = (3-1)\hat{i} + (-4 - (-3))\hat{j} + (-4 - (-5))\hat{k} = 2\hat{i} - \hat{j} + \hat{k}$$\vec{CA} = (2-3)\hat{i} + (-1 - (-4))\hat{j} + (1 - (-4))\hat{k} = -\hat{i} + 3\hat{j} + 5\hat{k}$Now,calculate the lengths of the sides:$|\vec{AB}|^2 = (-1)^2 + (-2)^2 + (-6)^2 = 1 + 4 + 36 = 41$$|\vec{BC}|^2 = (2)^2 + (-1)^2 + (1)^2 = 4 + 1 + 1 = 6$$|\vec{CA}|^2 = (-1)^2 + (3)^2 + (5)^2 = 1 + 9 + 25 = 35$Since $|\vec{AB}|^2 = |\vec{BC}|^2 + |\vec{CA}|^2$ $(41 = 6 + 35)$,the triangle satisfies the Pythagorean theorem.Therefore,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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