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

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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 $C(3, -4, -4)$.The side vectors are:$\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}$The lengths of the sides are:$c = |\vec{AB}| = \sqrt{(-1)^2 + (-2)^2 + (-6)^2} = \sqrt{1 + 4 + 36} = \sqrt{41}$$a = |\vec{BC}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$$b = |\vec{CA}| = \sqrt{(-1)^2 + 3^2 + 5^2} = \sqrt{1 + 9 + 25} = \sqrt{35}$Checking for the right-angled triangle condition $(a^2 + b^2 = c^2)$:$a^2 + b^2 = 6 + 35 = 41$$c^2 = 41$Since $a^2 + b^2 = c^2$,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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