[asy] import three; currentprojection = perspective(5,-40,12); defaultpen(linewidth(0.7)); pen l = linewidth(0.5) + linetype("10 2");  triple S=(1,0,0), T=(2,0,2), U=(8,6,8), V=(8,8,6), W=(2,2,0), X=(6,8,8); draw((1,0,0)--(8,0,0)--(8,0,8)--(0,0,8)--(0,0,1)); draw((1,0,0)--(8,0,0)--(8,8,0)--(0,8,0)--(0,1,0),l); draw((0,8,0)--(0,8,8)); draw((0,8,8)--(0,0,8)--(0,0,1)); draw((8,8,0)--(8,8,6),l); draw((8,0,8)--(8,6,8)); draw((0,8,8)--(6,8,8));  draw(S--T--U--V--W--cycle); draw((0,0,1)--T--U--X--(0,2,2)--cycle); draw((0,1,0)--W--V--X--(0,2,2)--cycle); [/asy]           [asy] import three; currentprojection = perspective(5,40,12); defaultpen(linewidth(0.7)); pen l = linewidth(0.5) + linetype("10 2");  triple S=(1,0,0), T=(2,0,2), U=(8,6,8), V=(8,8,6), W=(2,2,0), X=(6,8,8); draw((1,0,0)--(8,0,0)--(8,0,8),l); draw((8,0,8)--(0,0,8)); draw((0,0,8)--(0,0,1),l); draw((8,0,0)--(8,8,0)); draw((8,8,0)--(0,8,0)); draw((0,8,0)--(0,1,0),l); draw((0,8,0)--(0,8,8)); draw((0,0,8)--(0,0,1),l); draw((8,8,0)--(8,8,6)); draw((8,0,8)--(8,6,8)); draw((0,0,8)--(0,8,8)--(6,8,8));  draw(S--T--U--V--W--cycle); draw((0,0,1)--T--U--X--(0,2,2)--cycle); draw((0,1,0)--W--V--X--(0,2,2)--cycle); [/asy]
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, 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 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 $372 + 39 + 6 = 417$.
[asy] import three; currentprojection = perspective(5,-40,12); defaultpen(linewidth(0.7)); pen l = linewidth(0.5) + linetype("10 2"); triple S=(1,0,0), T=(2,0,2), U=(8,6,8), V=(8,8,6), W=(2,2,0), X=(6,8,8); draw((1,0,0)-(8,0,0)--(8,0,8)-(0,0,8)-(0,0,8)-(0,0,1)); draw((1,0,0)--(8,0,0)--(8,8,0)--(0,8,0)--(0,1,0),l); draw((0,8,0)--(0,8,8)); draw((0,8,8)--(0,0,8)--(0,0,1)); draw((8,8,0)--(8,8,6),l); draw((8,0,8)--(8,6,8)); draw((0,8,8)--(6,8,8)); draw(S--T--U--V--W--cycle); draw((0,0,1)--T--U--X--(0,2,2)--cycle); draw((0,1,0)--W--V--X--(0,2,2)--cycle); [/asy] [asy] import three; currentprojection = perspective(5,40,12); defaultpen(linewidth(0.7)); pen l = linewidth(0.5) + linetype("10 2"); triple S=(1,0,0), T=(2,0,2), U=(8,6,8), V=(8,8,6), W=(2,2,0), X=(6,8,8); draw((1,0,0)--(8,0,0)--(8,0,8),l); draw((8,0,8)--(0,0,8)); draw((0,0,8)--(0,0,1),l); draw((8,0,0)--(8,8,0)); draw((8,8,0)--(0,8,0)); draw((0,8,0)--(0,1,0),l); draw((0,8,0)--(0,8,8)); draw((0,0,8)--(0,0,1),l); draw((8,8,0)--(8,8,6)); draw((8,0,8)--(8,6,8)); draw((0,0,8)--(0,8,8)--(6,8,8));  draw(S--T--U--V--W--cycle); draw((0,0,1)--T--U--X--(0,2,2)--cycle); draw((0,1,0)--W--V--X--(0,2,2)--cycle); [/asy]
드릴링이 시작되는 정점 $E$가 $(8,8,8)$에 위치하도록 좌표계를 설정합니다. 약간의 시각화(평행선이 있으므로 유사한 삼각형 몇 개 포함)를 사용하면 터널이 $(1,0,0)에서 $(2,2,0)$, $(0,1,0)에서 $(2,2,0)$로 이어지는 선분에서 아래쪽 면(xy 평면의 면)과 만나고 다른 세 면도 마찬가지로 (대칭으로) 원점에서 만나는 것을 볼 수 있습니다. 따라서 터널의 한 면은 꼭지점이 $S(1,0,0), T(2,0,2), U(8,6,8), V(8,8,6), W(2,2,0)$인 다각형이고, 터널의 다른 두 면은 이 모양과 일치합니다.
이 도형은 높이가 $\sqrt {2}$이고 밑변이 $7\sqrt {3}$과 $6\sqrt {3}$인 두 개의 합동 사다리꼴로 구성되어 있음을 관찰합니다. 이 둘을 합하면 $\sqrt {2}(7\sqrt {3} + 6\sqrt {3}) = 13\sqrt {6}$의 면적을 구성합니다. 그러면 터널의 총 면적은 $3\cdot13\sqrt {6} = 39\sqrt {6}$입니다. 모퉁이 $E$ 주변에는 $6$의 면적이 누락되어 있으며, 반대쪽 모퉁이 $E$ 도 마찬가지입니다. 따라서 외부 면적은 $6\cdot 64 - 2\cdot 6 = 372$ 입니다. 따라서 총 표면적은 $372 + 39\sqrt {6}$이고, 답은 $372 + 39 + 6 = 417$입니다.