Consider the [set](https://artofproblemsolving.com/wiki/index.php/Set) of [points](https://artofproblemsolving.com/wiki/index.php/Point) that are inside or within one unit of a [rectangular parallelepiped](https://artofproblemsolving.com/wiki/index.php/Rectangular_prism) (box) that measures $3$ by $4$ by $5$ units. Given that the [volume](https://artofproblemsolving.com/wiki/index.php/Volume) of this set is $\frac{m + n\pi}{p},$ where $m, n,$ and $p$ are positive [integers](https://artofproblemsolving.com/wiki/index.php/Integer), and $n$ and $p$ are [relatively prime](https://artofproblemsolving.com/wiki/index.php/Relatively_prime), find $m + n + p.$
The set can be broken into several parts: the big $3\times 4 \times 5$ parallelepiped, $6$ external parallelepipeds that each share a face with the large parallelepiped and have a height of $1$, the $1/8$ [spheres](https://artofproblemsolving.com/wiki/index.php/Sphere) (one centered at each [vertex](https://artofproblemsolving.com/wiki/index.php/Vertex) of the large parallelepiped), and the $1/4$ [cylinders](https://artofproblemsolving.com/wiki/index.php/Cylinder) connecting each adjacent pair of spheres.

The volume of the parallelepiped is $3 \times 4 \times 5 = 60$ cubic units.
The volume of the external parallelepipeds is $2(3 \times 4 \times 1)+2(3 \times 5 \times 1 )+2(4 \times 5 \times 1)=94$.
There are $8$ of the $1/8$ spheres, each of radius $1$. Together, their volume is $\frac{4}{3}\pi$.
There are $12$ of the $1/4$ cylinders, so $3$ complete cylinders can be formed. Their volumes are $3\pi$, $4\pi$, and $5\pi$, adding up to $12\pi$.
The combined volume of these parts is $60+94+\frac{4}{3}\pi+12\pi = \frac{462+40\pi}{3}$. Thus, the answer is $m+n+p = 462+40+3 = \boxed{505}$.