Example 11 Rationalize the denominator:
$$
\frac{3+2 \sqrt{2}-\sqrt{3}-\sqrt{6}}{1+\sqrt{2}-\sqrt{3}}=
$$
$\qquad$
(Fifth National Partial Provinces and Cities Junior High School Mathematics Competition)
Solution: From $\sqrt{6}=\sqrt{2} \times \sqrt{3}$, we know that the numerator of the original expression is the product of two radicals containing $\sqrt{2}$ and $\sqrt{3}$. Therefore, we can set
$$
\begin{aligned}
& 3+2 \sqrt{2}-\sqrt{3}-\sqrt{6} \\
& =(1+\sqrt{2}-\sqrt{3})(a+b \sqrt{2}+c \sqrt{3}) \\
= & (a+2 b-3 c)+(a+b) \sqrt{2} \\
+ & (c-a) \sqrt{3}+(c-b) \sqrt{6} .
\end{aligned}
$$

By comparing the coefficients of the corresponding terms on both sides of the equation, we get
$$
\left\{\begin{array}{l}
a+2 b-3 c=3, \\
a+b=2, \\
c-a=-1, \\
c-b=-1 .
\end{array}\right.
$$

Solving these equations, we find $a=1, b=1, c=0$.
$$
\begin{aligned}
\therefore & \frac{3+2 \sqrt{2}-\sqrt{3}-\sqrt{0}}{1+\sqrt{2}-\sqrt{3}} \\
& =\frac{(1+\sqrt{2}-\sqrt{3})(1+\sqrt{2})}{1+\sqrt{2}-\sqrt{3}}=1+\sqrt{2} .
\end{aligned}
$$