3. Given $f(x)=x^{2}+6 a x-a, y=f(x)$, the graph of which intersects the $x$-axis at two distinct points $\left(x_{1}, 0\right),\left(x_{2}\right.$, $0)$, and
$$
\begin{array}{l}
\frac{a}{\left(1+x_{1}\right)\left(1+x_{2}\right)}-\frac{3}{\left(1-6 a-x_{1}\right)\left(1-6 a-x_{2}\right)} \\
=8 a-3 .
\end{array}
$$

Then the value of $a$ is ( ).
(A) 1
(B) 2
(C) 0 or $\frac{1}{2}$
(D) $\frac{1}{2}$
3. D.

From $\Delta=36 a^{2}+4 a>0$, we get $a>0$ or $a<-\frac{1}{9}$. By the problem, we can set
$$
\begin{array}{l}
f(x)=x^{2}+6 a x-a=\left(x-x_{1}\right)\left(x-x_{2}\right) . \\
\text { Then }\left(1+x_{1}\right)\left(1+x_{2}\right)=f(-1)=1-7 a, \\
\left(1-6 a-x_{1}\right)\left(1-6 a-x_{2}\right) \\
=f(1-6 a)=1-7 a .
\end{array}
$$

Therefore, $\frac{a-3}{1-7 a}=8 a-3$.
Solving this, we get $a=\frac{1}{2}$ or $a=0$ (discard).