3. Let $[x]$ denote the greatest integer not exceeding the real number $x$. If the real number $a$ satisfies
$$
a-\frac{4}{a}+\frac{3 \sqrt{a(a-2)}}{a}=2,
$$
then $[a]=$ ( ).
(A) 0 or 2
(B) -1 or 2
(C) 0 or 3
(D) 2 or -3
3. B.

Equation (1) $\Leftrightarrow a^{2}-4+3 \sqrt{a(a-2)}=2 a$. Let $\sqrt{a(a-2)}=x$. Then $x^{2}+3 x-4=0$. Solving, we get $x_{1}=1, x_{2}=-4$ (discard). Thus, $\sqrt{a(a-2)}=1$.
Rearranging gives $a^{2}-2 a-1=0 \Rightarrow a_{1,2}=1 \pm \sqrt{2}$. Therefore, $[a]=-1$ or 2.