Example 2 Given $m=1+\sqrt{2}, n=1-\sqrt{2}$, and $\left(7 m^{2}-14 m+a\right)\left(3 n^{2}-6 n-7\right)=8$. Then the value of $a$ is ( ).
$(\mathrm{A})-5$
(B) 5
(C) -9
(D) 9
Given that $m+n=2, m n=-1$.
Therefore, $m, n$ can be seen as the roots of the quadratic equation $x^{2}-2 x-1=0$.
Thus, $m^{2}-2 m=1, n^{2}-2 n=1$.
Since $\left(7 m^{2}-14 m+a\right)\left(3 n^{2}-6 n-7\right)=8$,
So, $(7+a)(3-7)=8$.
Solving for $a$ gives $a=-9$.
Therefore, the answer is C.