Given, $ 3 p +2 q = i + j + k\,\,\,\,\,\,\,\dots(i)$
and $ 3 p-2 q=i-j-k\,\,\,\,\,\,\dots(ii)$
On adding Eqs. (i) and (ii), we get
$6p =2 i$
$\therefore \, p=\frac{i}{3}$
On subtracting Eq. (ii) from Eq. (i), we get
$4 q =2( j + k ) $
$\Rightarrow \, q =\frac{1}{2}( j + k )$
Let $\theta$ be the angle between $p$ and $q$, then
$ p \cdot q =| p || q | \cos \,\theta$
$\Rightarrow \, \cos \theta=\frac{ p \cdot q }{| p || q |}$
$\Rightarrow \, \cos \theta=\frac{1}{6} \frac{ i \cdot( j + k )}{| p || q |}=0$
$=\cos 90^{\circ}$
$\therefore \, \theta=90^{\circ}$
$=\frac{\pi}{2}$