Show that the vectors $2\hat i - \hat j + \hat k,\hat i - 3\hat j - 5\hat k$ and $3\hat i - 4\hat j - 4\hat k$ form the vertices of a right angled triangle.

Let the position vectors of vertices $A,B$ and
$C$ be $2\hat i - \hat j + \hat k,\hat i - 3\hat j - 5\hat k$ and $3\hat i - 4\hat j - 4\hat k$ respectively.

$\therefore$ $\overrightarrow {AB} = \overrightarrow {OB} - \overrightarrow {OA} = (\hat i - 3\hat j - 5\hat k) - (2\hat i - \hat j + \hat k) = - \hat i - 2\hat j - 6\hat k$

$\therefore$ $AB = |\overrightarrow {AB} | = \sqrt {{{( - 1)}^2} + {{( - 2)}^2} + {{( - 6)}^2}} = \sqrt {1 + 4 + 36} = \sqrt {41}$

$\overrightarrow {BC} = \overrightarrow {OC} - \overrightarrow {OB} = (3\hat i - 4\hat j - 4\hat k) - (\hat i - 3\hat j - 5\hat k) = 2\hat i - \hat j + \hat k$

$\therefore$ $BC = |\overrightarrow {BC} | = \sqrt {{2^2} + {{( - 1)}^2} + {1^2}} = \sqrt {4 + 1 + 1} = \sqrt 6$and

$\overrightarrow {AC} = \overrightarrow {OC} - \overrightarrow {OA} = (3i - 4\hat j - 4\hat k) - (2\hat i - \hat j + \hat k) = \hat i - 3\hat j - 5\hat k$

$\therefore$ $AC = |\overrightarrow {AC} | = \sqrt {{1^2} + {{( - 3)}^2} + {{( - 5)}^2}} = \sqrt {1 + 9 + 25} = \sqrt {35}$

Now, $B{C^2} + A{C^2} = 6 + 35 = 41 = A{B^2}$
Hence, $\Delta ABC$ is a right angled triangle.