3. If the equation $\left(\frac{3}{2}\right)^{x}=\frac{2+3 a}{5-a}$ has a negative root for $x$, then the range of real number $a$ is ( ).
(A) $\left(-\infty,-\frac{2}{3}\right) \cup(5,+\infty)$
(B) $\left(-\infty,-\frac{3}{4}\right) \cup(5,+\infty)$
(C) $\left(-\frac{2}{3}, 5\right)$
(D) $\left(-\frac{2}{3}, \frac{3}{4}\right)$
3. D.

Since $x<0$, we have $0<\left(\frac{3}{2}\right)^{x}<1$. Therefore, $0<\frac{2+3a}{5-a}<1$. Solving this, we get $-\frac{2}{3}<a<\frac{3}{4}$.