3. If $B$ is a point on the diameter $AC$ of a circle with radius 3 and $\angle ABC = 90^\circ$, $BC = 2$, and circles $\odot A, \odot C, \odot P$ are constructed with $AB, BC$ as radii, respectively, such that $\odot P$ is tangent to $\odot A, \odot C, \odot O$, then the radius $r$ of $\odot P$ is $\qquad$
3. $\frac{2}{3}$.

As shown in Figure 6, connect $P A$, $P O$, $P B$, we get $P A=4+r$. $O P=3-r$, $P B=2+r$. Let $\angle A O P=\alpha, \angle B O P=\beta$, then
$$
\begin{array}{l}
\alpha+\beta=180^{\circ}, \\
\cos \alpha=-\cos \beta .
\end{array}
$$

By the Law of Cosines, we have
$$
\begin{array}{l}
\frac{3^{2}+(3-r)^{2}-(4+r)^{2}}{2 \cdot 3 \cdot(3-r)} \\
=-\frac{3^{2}+(3-r)^{2}-(2+r)^{2}}{2 \cdot 3 \cdot(3-r)} .
\end{array}
$$

Simplifying, we get $r=\frac{2}{3}$.