If  $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=a+b\sqrt{6}$, then $(a,b)=$

$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=a+b\sqrt{6}$

R.F of  $\sqrt{3}-\sqrt{2}$ is $\sqrt{3}+\sqrt{2}$

$\frac{(\sqrt{3}+\sqrt{2})(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}=a+b\sqrt{6}$

$\frac{{(\sqrt{3}+\sqrt{2})}^2}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2}=a+b\sqrt{6}$

$\frac{3+2+2\sqrt{6}}{3-2}=a+b\sqrt{6}$

$\begin{array}{l}5+2\sqrt{6}=a+b\sqrt{6}\\ a=5,b=2\\ (a,b)=(5,2)\end{array}$