Given, $a =2 i +3 j +\alpha k$ and $b =3 i -\alpha j +2 k$
$\therefore a + b =5 i +(3-\alpha) j +(\alpha+2) k$
and $a - b =- i +(3+\alpha) j +(\alpha-2) k$
$\therefore \cos \theta=\frac{( a + b ) \cdot( a - b )}{| a + b || a - b |}$
$=\frac{\left[\{5 i +(3-\alpha) j +(\alpha+2) k \} {- i +(3+\alpha) j +(\alpha-2) k }.\right]}{\left[\sqrt{5^{2}+(3-\alpha)^{2}+(\alpha+2)^{2}} \times \sqrt{1^{2}+(3+\alpha)^{2}+(\alpha-2)^{2}}\right]}$
$=\frac{-5+9-\alpha^{2}+\alpha^{2}-4}{\left[\sqrt{5^{2}+(3-\alpha)^{2}+(\alpha+2)^{2}} \left.\times \sqrt{1+(3+\alpha)^{2}+(\alpha-2)^{2}}\right]\right.}=0$
$\Rightarrow \theta=\frac{\pi}{2}$