Answer: 113

Small Hint:

Folding preserves distances: if the crease meets $\overline{AB}$ at $P,$ then $P$ is equidistant from $A$ and from the landing point on $\overline{BC}$

Big Hint:

The law of cosines in the $60^\circ$ corner triangles at $B$ and $C$ locates both crease endpoints; then apply the law of cosines once more at $A$

Solution:

Let $A'$ be the landing point, with $BA' = 9$ and $CA' = 3,$ and let the crease meet $\overline{AB}$ at $P$ and $\overline{AC}$ at $Q.$ Folding preserves distances, so $PA' = PA = x$ and $QA' = QA = y.$ In triangle $PBA',$ with $PB = 12 - x$ and $\angle B = 60^\circ,$ the law of cosines gives $$\begin{aligned} x^2 &= (12 - x)^2 + 81 \\ &\quad {}- 9(12 - x), \end{aligned}$$ which simplifies to $15x = 117,$ so $x = \frac{39}{5}.$ Similarly, in triangle $QCA',$ $y^2 = (12 - y)^2 + 9 - 3(12 - y)$ gives $21y = 117,$ so $y = \frac{39}{7}.$

Finally, in triangle $APQ$ with $\angle A = 60^\circ,$ $$\begin{aligned} PQ^2 &= x^2 + y^2 - xy \\ &= 39^2\left(\frac{1}{25} + \frac{1}{49} - \frac{1}{35}\right) \\ &= 39^2 \cdot \frac{49 + 25 - 35}{1225} \\ &= \frac{39^3}{1225}, \end{aligned}$$ so $PQ = \frac{39\sqrt{39}}{35}.$ Thus $m + n + p = 39 + 35 + 39$ $= 113.$