3. Given a triangle $\mathrm{PEF}$ with sides $\mathrm{PE}=3, \mathrm{PF}=5, \mathrm{EF}=7$. On the extension of side $\mathrm{FP}$ beyond point $\mathrm{P}$, a segment $\mathrm{PA}=1.5$ is laid out. Find the distance $d$ between the centers of the circumcircles of triangles $\mathrm{EPA}$ and $\mathrm{EAF}$. In the answer, specify the number equal to $2d$.
3. By the cosine theorem for angle EPF, we find that $\operatorname{Cos} E P F=-\frac{1}{2}$. Therefore, the angle $\mathrm{EPF}=120^{\circ}$; hence the adjacent angle will be $60^{\circ}$. Draw a perpendicular from point E to PF, meeting at point D. Since angle EPF is obtuse, point D will lie outside triangle EPF. In the right triangle PED, PD lies opposite the angle $30^{\circ}$, so $\mathrm{PD}=\frac{1}{2} \mathrm{EP}=1.5$. Therefore, point D coincides with point A. Thus, we have two right triangles: EAP and EAF. The centers of the circumcircles of these triangles lie on the midpoints of the corresponding hypotenuses. Therefore, the desired distance is equal to the midline of triangle EPF, i.e., $d=2.5$. Hence, the answer is: $\{5\}$.