Find the values of 'a' and 'b':

$\dfrac{3}{\sqrt{3} - \sqrt{2}} = a\sqrt{3} - b\sqrt{2}$

Given,

Equation : $\dfrac{3}{\sqrt{3} - \sqrt{2}} = a\sqrt{3} - b\sqrt{2}$

Rationalizing L.H.S. of the above equation :

$\Rightarrow \dfrac{3}{\sqrt{3} - \sqrt{2}} \times \dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} \\[1em] \Rightarrow \dfrac{3(\sqrt{3} + \sqrt{2})}{(\sqrt{3})^2 - (\sqrt{2})^2} \\[1em] \Rightarrow \dfrac{3\sqrt{3} + 3\sqrt{2}}{3 - 2} \\[1em] \Rightarrow \dfrac{3\sqrt{3} + 3\sqrt{2}}{1} \\[1em] \Rightarrow 3\sqrt{3} + 3\sqrt{2}.$

Comparing $3\sqrt{3} + 3\sqrt{2}\text{ with } a\sqrt{3} - b\sqrt{2}$, we get :

a = 3 and b = -3.

Hence, a = 3 and b = -3.