3. Let $[x]$ denote the greatest integer not exceeding the real number $x$ (for example, $[2.1]=2,[1]=1)$. If the real number $a$ satisfies
$$
a-\frac{5}{a}+\frac{4 \sqrt{a(a-3)}}{a}=3 \text {, }
$$

then $[a]=$ $\qquad$ .
3. -1 or 3.

The original equation is equivalent to
$a^{2}-5+4 \sqrt{a(a-3)}=3 a (a<0$ or $a \geqslant 3)$.
Let $x=\sqrt{a(a-3)} \geqslant 0$. Then
$x^{2}+4 x-5=0 \Rightarrow x_{1}=1, x_{2}=-5$ (discard).
Thus $\sqrt{a(a-3)}=1 \Rightarrow a^{2}-3 a-1=0$
$\Rightarrow a_{1}=\frac{3+\sqrt{13}}{2}, a_{2}=\frac{3-\sqrt{13}}{2}$
$\Rightarrow[a]=-1$ or 3.