Let $EFGH$, $EFDC$, and $EHBC$ be three adjacent [square](https://artofproblemsolving.com/wiki/index.php/Square) faces of a [cube](https://artofproblemsolving.com/wiki/index.php/Cube), for which $EC = 8$, and let $A$ be the eighth [vertex](https://artofproblemsolving.com/wiki/index.php/Vertex) of the cube. Let $I$, $J$, and $K$, be the points on $\overline{EF}$, $\overline{EH}$, and $\overline{EC}$, respectively, so that $EI = EJ = EK = 2$. A solid $S$ is obtained by drilling a tunnel through the cube. The sides of the tunnel are planes parallel to $\overline{AE}$, and containing the edges, $\overline{IJ}$, $\overline{JK}$, and $\overline{KI}$. The [surface area](https://artofproblemsolving.com/wiki/index.php/Surface_area) of $S$, including the walls of the tunnel, is $m + n\sqrt {p}$, where $m$, $n$, and $p$ are positive integers and $p$ is not divisible by the square of any prime. Find $m + n + p$.
Set the coordinate system so that vertex $E$, where the drilling starts, is at $(8,8,8)$. Using a little visualization (involving some [similar triangles](https://artofproblemsolving.com/wiki/index.php/Similar_triangles), because we have parallel lines) shows that the tunnel meets the bottom face (the xy plane one) in the line segments joining $(1,0,0)$ to $(2,2,0)$, and $(0,1,0)$ to $(2,2,0)$, and similarly for the other three faces meeting at the origin (by symmetry). So one face of the tunnel is the polygon with vertices (in that order), $S(1,0,0), T(2,0,2), U(8,6,8), V(8,8,6), W(2,2,0)$, and the other two faces of the tunnel are congruent to this shape. 
Observe that this shape is made up of two congruent [trapezoids](https://artofproblemsolving.com/wiki/index.php/Trapezoid) each with height $\sqrt {2}$ and bases $7\sqrt {3}$ and $6\sqrt {3}$. Together they make up an area of $\sqrt {2}(7\sqrt {3} + 6\sqrt {3}) = 13\sqrt {6}$. The total area of the tunnel is then $3\cdot13\sqrt {6} = 39\sqrt {6}$. Around the corner $E$ we're missing an area of $6$, the same goes for the corner opposite $E$ . So the outside area is $6\cdot 64 - 2\cdot 6 = 372$. Thus the the total surface area is $372 + 39\sqrt {6}$, and the answer is $372 + 39 + 6 = \boxed{417}$.