3. Given the equation in terms of $x$
$$
a x^{2}+(a+1) x+6 a=0
$$

has two distinct real roots $x_{1} 、 x_{2}\left(x_{1}<1<x_{2}\right)$. Then the range of the real number $a$ is ( ).
(A) $-1<a<0$
(B) $a<-1$
(C) $-\frac{1}{8}<a<0$
(D) $a<-\frac{1}{8}$
3. C.

From the given information, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
a \neq 0, \\
\Delta=(a+1)^{2}-4 a \times 6 a>0, \\
x_{1}+x_{2}=-\frac{a+1}{a}>0, \\
x_{1} x_{2}=\frac{6 a}{a}=6
\end{array}\right. \\
\Rightarrow-\frac{1}{8}<a<0 .
\end{array}
$$