Show that the points $A (2 \hat{i}-\hat{j}+\hat{k})$,$B (\hat{i}-3 \hat{j}-5 \hat{k})$,and $C (3 \hat{i}-4 \hat{j}-4 \hat{k})$ are the vertices of a right-angled triangle.

(A) The position vectors of the vertices are $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$,$\vec{b} = \hat{i} - 3\hat{j} - 5\hat{k}$,and $\vec{c} = 3\hat{i} - 4\hat{j} - 4\hat{k}$.
First,we find the vectors representing the sides:
$\vec{AB} = \vec{b} - \vec{a} = (1-2)\hat{i} + (-3+1)\hat{j} + (-5-1)\hat{k} = -\hat{i} - 2\hat{j} - 6\hat{k}$
$\vec{BC} = \vec{c} - \vec{b} = (3-1)\hat{i} + (-4+3)\hat{j} + (-4+5)\hat{k} = 2\hat{i} - \hat{j} + \hat{k}$
$\vec{CA} = \vec{a} - \vec{c} = (2-3)\hat{i} + (-1+4)\hat{j} + (1+4)\hat{k} = -\hat{i} + 3\hat{j} + 5\hat{k}$
Now,calculate the squares of the magnitudes of these sides:
$|\vec{AB}|^2 = (-1)^2 + (-2)^2 + (-6)^2 = 1 + 4 + 36 = 41$
$|\vec{BC}|^2 = (2)^2 + (-1)^2 + (1)^2 = 4 + 1 + 1 = 6$
$|\vec{CA}|^2 = (-1)^2 + (3)^2 + (5)^2 = 1 + 9 + 25 = 35$
Observe that $|\vec{AB}|^2 = |\vec{BC}|^2 + |\vec{CA}|^2$ since $41 = 6 + 35$.
By the converse of the Pythagoras theorem,the triangle is a right-angled triangle with the right angle at vertex $C$.