3. Given that $f(x)$ is an odd function with a period of 4, and when $x \in(0,2)$, $f(x)=x^{2}-16 x+60$, then the value of $f(2 \sqrt{10})$ is $\qquad$ .
$$
6<2 \sqrt{10}<7 \Rightarrow f(2 \sqrt{10})=f(2 \sqrt{10}-8)=-f(8-2 \sqrt{10}) \text {, }
$$

and $x=8-2 \sqrt{10} \Rightarrow x-8=-2 \sqrt{10} \Rightarrow x^{2}-16 x+64=40 \Rightarrow x^{2}-16 x+60=36 \Rightarrow f(2 \sqrt{10})=-36$.