3. Let the inradius of $\triangle A B C$ be $r, B C=a$, $A C=b, A B=c$, and the altitudes be $h_{a} 、 h_{b}$ 、 $h_{c}$, satisfying $h_{a}+h_{b}+h_{c}=9 r$. Then the shape of $\triangle A B C$ is ( ).
(A) must be an obtuse triangle
(B) must be an equilateral triangle
(C) must not be an acute triangle
(D) may not be a right triangle
3. B.

It is known that $S_{\triangle A B C}=\frac{1}{2} a h_{a}=\frac{1}{2} b h_{b}=\frac{1}{2} c h_{c}$, so we have $h_{a}+h_{b}+h_{c}=2 S_{\triangle I B C}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$.

Since $S_{\triangle A B C}=\frac{1}{2} r(a+b+c)$,
$$
h_{a}+h_{b}+h_{c}=9 r \text {, }
$$

then we have $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=9$, which means $a(b-c)^{2}+b(c-a)^{2}+c(a-b)^{2}=0$, so, $(a-b)^{2}=(b-c)^{2}=(c-a)^{2}=0$. Hence, $a=b=c$.