Answer : $\;-\large\frac{7\sqrt{2}}{2} \normalsize i$
Explanation :
$\large\frac{(\normalsize 3 + \normalsize i \sqrt{5})(\normalsize 3 - \normalsize i \sqrt{5})}{(\normalsize \sqrt{3} + \normalsize i \sqrt{2})-(\normalsize \sqrt{3} - \normalsize i \sqrt{2})}= $
$=\large\frac{(3)^{2}-(\sqrt{5})^{2}}{\sqrt{3}+\sqrt{2} i -\sqrt{3}+\sqrt{2}i}\qquad [(a+b)(a-b)=a^2-b^2]$
$= \large\frac{9-5i^{2}}{2\sqrt{2}i}$
$= \large\frac{9+5}{2 \sqrt{2} i} \times \large\frac{i}{i}\qquad [i^{2}=-1]$
$= \large\frac{14}{2 \sqrt{2} i^2}$
$ =- \large\frac{7i}{\sqrt{2}} \times \large\frac{\sqrt{2}}{\sqrt{2}}$
$=-\large\frac{7\sqrt{2}}{2} \normalsize i$