Find a unit vector perpendicular to each of the vectors $\vec a + \vec b$ and $\vec a - \vec b$, where $\vec a = 3\hat i + 2\hat j + 2\hat k$ and $\vec b = \hat i + 2\hat j - 2\hat k$.

We have, $\vec a = 3\hat i + 2\hat j + 2\hat k$ and $\vec b = \hat i + 2\hat j - 2\hat k$

$\therefore$ $\vec a + \vec b = (3\hat i + 2\hat j + 2\hat k) + (\hat i + 2\hat j - 2\hat k) = 4\hat i + 4\hat j$

and $\vec a - \vec b = (3\hat i + 2\hat j + 2\hat k) - (\hat i + 2\hat j - 2\hat k) = 2\hat i + 4\hat k$
$\therefore$ $(\vec a + \vec b) \times (\vec a - \vec b) = \left| {\begin{array}{llllllllllllllllllll}{\hat i}&{\hat j}&{\hat k}\\4&4&0\\2&0&4\end{array}} \right|$

$= (16 - 0)\hat i - (16 - 0)\hat j + (0 - 8)\hat k = 16\hat i - 16\hat j - 8\hat k$
$\therefore$

Unit vector perpendicular to both $(\vec a + \vec b)\,{\rm{and}}\,\,(\vec a - \vec b)$ is given by

$\hat n = \pm \cfrac{{(\vec a + \vec b) \times (\vec a - \vec b)}}{{|(\vec a + \vec b) \times (\vec a - \vec b)|}} = \pm \cfrac{{16\hat i - 16\hat j - 8\hat k}}{{\sqrt {{{(16)}^2} + {{( - 16)}^2} + {{( - 8)}^2}} }}$

$= \pm \cfrac{{8(2\hat i - 2\hat j - \hat k)}}{{8\sqrt {4 + 4 + 1} }} = \pm \cfrac{{2\hat i - 2\hat j - \hat k}}{3} = \pm \cfrac{2}{3}\hat i \mp \cfrac{2}{3}j \mp \cfrac{1}{3}\hat k$