Answer: 36

Small Hint:

In triangle $ABD,$ the bisector from $B$ meets the perpendicular bisector of $AD$ at the arc midpoint of $AD,$ so $\angle EAD = \frac{B}{2};$ similarly $\angle FAD = \frac{C}{2}.$

Big Hint:

With $M$ the midpoint of $\overline{AD},$ the area is $\frac{1}{2}AM^2\left(\tan\frac{B}{2} + \tan\frac{C}{2}\right),$ and $AD^2 = AB \cdot AC - BD \cdot DC.$

Solution:

In triangle $ABD,$ the internal bisector of the angle at $B$ meets the circumcircle of $ABD$ again at the midpoint of arc $AD$ not containing $B,$ and that arc midpoint lies on the perpendicular bisector of $\overline{AD}$ — so $E$ is exactly that arc midpoint. The inscribed angles $\angle EAD$ and $\angle EBD$ subtend the same arc $ED,$ so $\angle EAD = \frac{B}{2}.$ Similarly $\angle FAD = \frac{C}{2},$ and $E, F$ lie on opposite sides of line $AD.$

Let $M$ be the midpoint of $\overline{AD}.$ In right triangles $AME$ and $AMF,$ $ME = AM\tan\frac{B}{2}$ and $MF = AM\tan\frac{C}{2},$ so $EF = AM\left(\tan\frac{B}{2} + \tan\frac{C}{2}\right),$ while the distance from $A$ to line $EF$ is $AM.$ Hence $[AEF]$ $= \frac{1}{2}AM^2\left(\tan\frac{B}{2} + \tan\frac{C}{2}\right).$

Here $BD = 2$ and $DC = 3,$ so $AD^2 = AB \cdot AC - BD \cdot DC$ $= 24 - 6$ $= 18$ and $AM^2 = \frac{9}{2}.$ The law of cosines gives $\cos B = \frac{1}{8}$ and $\cos C = \frac{3}{4},$ so $\tan\frac{B}{2} = \sqrt{\frac{1 - \frac{1}{8}}{1 + \frac{1}{8}}} = \frac{\sqrt{7}}{3}$ and $\tan\frac{C}{2} = \frac{1}{\sqrt{7}},$ with sum $\frac{10\sqrt{7}}{21}.$ The area is $\frac{1}{2} \cdot \frac{9}{2} \cdot \frac{10\sqrt{7}}{21} = \frac{15\sqrt{7}}{14},$ so $m + n + p = 15 + 7 + 14 = 36.$