Example 4 Let $a=\sqrt{7}-1$. Then $3 a^{3}+12 a^{2}-6 a-12=(\quad)$.
(A) 24
(B) 25
(C) $4 \sqrt{7}+10$
(D) $4 \sqrt{7}+12$
Given $a=\sqrt{7}-1$, we get $a+1=\sqrt{7}$.
Thus, $a^{2}+2 a-6=0$.
Therefore, the expression to be found is
$$
\begin{array}{l}
=3 a^{3}+6 a^{2}+6 a^{2}-6 a-12 \\
=3 a\left(a^{2}+2 a\right)+6 a^{2}-6 a-12 \\
=18 a+6 a^{2}-6 a-12 \\
=6\left(a^{2}+2 a\right)-12 \\
=6 \times 6-12=24 .
\end{array}
$$

Therefore, the answer is A.