Example 12 Simplify the radical expression:
$$
\sqrt{2(6-2 \sqrt{3}-2 \sqrt{5}+\sqrt{15})} \text {. }
$$
(1997, Taiyuan City Junior High School Mathematics Competition)
Solution: Let $2(6-2 \sqrt{3}-2 \sqrt{5}+\sqrt{15})=(a+$ $b \sqrt{3}+c \sqrt{5})^{2}$, that is,
$$
\begin{aligned}
12 & -4 \sqrt{3}-4 \sqrt{5}+2 \sqrt{15} \\
= & \left(a^{2}+3 b^{2}+5 c^{2}\right)+2 a b \sqrt{3}+2 a c \sqrt{5} \\
& +2 b c \sqrt{15} .
\end{aligned}
$$

Comparing the coefficients of the corresponding terms on both sides of the equation, we get
$$
\left\{\begin{array}{l}
a^{2}+3 b^{2}+5 c^{2}=12, \\
2 a b=-4, \\
2 a c=-4, \\
2 b c=2 .
\end{array}\right.
$$

Solving these, we get $a=2, b=-1, c=-1$ or $a=-2$, $b=1, c=1$.
$$
\begin{aligned}
\therefore & \sqrt{2(6-2 \sqrt{3}-2 \sqrt{5}+\sqrt{15})} \\
= & \sqrt{( \pm 2 \mp \sqrt{3} \mp \sqrt{5})^{2}} \\
= & | \pm 2 \mp \sqrt{3} \mp \sqrt{5}| \\
= & -2+\sqrt{3}+\sqrt{5} .
\end{aligned}
$$