$13 \cdot 27$ In the three-dimensional Cartesian coordinate system, $E$ is the set of triangles with vertices at integer points, and no other integer points inside or on the edges. Find the set of areas of triangles, $f(E)$.
(30th International Mathematical Olympiad Candidate Problem, 1989)
[Solution] Let $\triangle OAB$ be a triangle in $E$, where $O$ is the origin.
It is easy to know that the equation of the plane $OAB$ is
$$
\left|\begin{array}{lll}
x & x_{A} & x_{B} \\
y & y_{A} & y_{B} \\
z & z_{A} & z_{B}
\end{array}\right|=0
$$

where $\left(x_{A}, y_{A}, z_{A}\right),\left(x_{B}, y_{B}, z_{B}\right)$ are the coordinates of $A, B$ respectively.
Expanding the determinant and canceling the common factor, we can assume the equation of the plane $AOB$ is
$$
a x + b y + c z = 0,
$$

where $a, b, c$ are integers, and the greatest common divisor of $a, b, c$ is 1.
Since the integer point $(x, y, z)$ makes $a x + b y + c z$ an integer, there are obviously no integer points between the parallel planes $a x + b y + c z = 0$ and $a x + b y + c z = 1$ (excluding these two planes).
Since the greatest common divisor of $a, b, c$ is 1, there exists a set of integers $\left(x_{c}, y_{c}, z_{c}\right)$ such that
$$
a x_{c} + b y_{c} + c z_{c} = 1
$$

holds.
That is, the integer point $C\left(x_{c}, y_{c}, z_{c}\right)$ lies on the plane $a x + b y + c z = 1$.
A parallelepiped can be constructed with $OA, OB, OC$ as edges. Since there are no integer points inside or on the edges of $\triangle OAB$ (excluding the vertices), the parallelogram with $OA, OB$ as sides is also such. Therefore, the constructed parallelepiped has no other integer points inside or on its faces (excluding the vertices).

Starting from this basic parallelepiped, a hexahedral network can be formed in space, where each integer point is a vertex of some hexahedron, but will not appear inside the hexahedron (or on the faces or edges of the hexahedron).
That is, for any set of integers $(x, y, z)$, the system of equations
$$
\left\{\begin{array}{l}
l x_{A} + m x_{B} + n x_{c} = x, \\
l y_{A} + m y_{B} + n y_{c} = y, \\
l z_{A} + m z_{B} + n z_{c} = z .
\end{array}\right.
$$

has integer solutions $(l, m, n)$, so it must be that
$$
\left|\begin{array}{lll}
x_{A} & x_{B} & x_{C} \\
y_{A} & y_{B} & y_{C} \\
z_{A} & z_{B} & z_{C}
\end{array}\right| = \pm 1 .
$$

That is, the volume of the basic parallelepiped is 1.
Since the distance between the planes $a x + b y + c z = 0$ and $a x + b y + c z = 1$ is
$$
\frac{1}{\sqrt{a^{2} + b^{2} + c^{2}}},
$$

the area of $\triangle OAB$ is
$$
\frac{1}{2} \sqrt{a^{2} + b^{2} + c^{2}}.
$$

Thus, $f(E) = \left\{\frac{1}{2} \sqrt{a^{2} + b^{2} + c^{2}}\right\} a, b, c \in \mathbb{Z}, \quad (a, b, c) = 1\}$.