Simplify by rationalizing the denominator $\frac{2\sqrt{3}-\sqrt{2}}{2\sqrt{2}+3\sqrt{3}}$

The correct option is B: $\frac{22-7\sqrt{6}}{19}$
$\frac{2\sqrt{3}-\sqrt{2}}{2\sqrt{2}+3\sqrt{3}}$

the conjugate of $2\sqrt{2}+3\sqrt{3}$ is $2\sqrt{2}-3\sqrt{3}$

$=\frac{(2\sqrt{3}-\sqrt{2})}{(2\sqrt{2}+3\sqrt{3})}\times$ $\frac{(2\sqrt{2}-3\sqrt{3})}{(2\sqrt{2}-3\sqrt{3})}$

$=\frac{(2\sqrt{3}\times 2\sqrt{2})+(2\sqrt{3}\times -3\sqrt{3})+(-\sqrt{2}\times 2\sqrt{2})+(-\sqrt{2}\times -3\sqrt{3})}{(2\sqrt{2}{)}^2-(3\sqrt{3}{)}^2}$ $\dots \left[\right(a+b\left)\right(a-b)=a^2-b^2]$

$=\frac{4\sqrt{6}-6\sqrt{9}-2\sqrt{4}+3\sqrt{6}}{8-27}$ [$\sqrt{4}=2,\sqrt{9}=3$]

$=\frac{4\sqrt{6}-18-4+3\sqrt{6}}{-19}$

$=\frac{7\sqrt{6}-22}{-19}$

$=\frac{22-7\sqrt{6}}{19}$