$9 \cdot 74$ Let the length, width, and height of a rectangular prism be positive integers $m, n, r$ and $m \leqslant n \leqslant r$. After painting the surface of the rectangular prism red, it is cut into unit cubes. It is known that the number of unit cubes with no red faces plus the number of unit cubes with two red faces, minus the number of unit cubes with one red face equals 1985. Find the values of $m, n$, and $r$.
[Solution] Let the number of cubes with no red, one red face, and two red faces be denoted as $k_{0}, k_{1}$, and $k_{2}$, respectively.
(1) Let $m=1$. Then $k_{0}=k_{1}=0$. When $n \leqslant 2$, $k_{2}=0$, which contradicts $k_{0}-k_{1}+k_{2}=1985$. Therefore, we know that $n>2$. At this time, we have
$$
k_{2}=(n-2)(r-2)=1985 \text {. }
$$

Since $1985=1 \times 1985=5 \times 397$, and 397 is a prime number, we get two sets of solutions $(1,3, 1987)$, $(1,7,399)$.
(2) Let $m=2$. At this time, $k_{0}=0$. By the symmetry of the cuboid, we know that $k_{1}$ and $k_{2}$ are both even, and their difference cannot be 1985, which is a contradiction. Therefore, there is no solution in this case.
(3) Let $m \geqslant 3$. At this time, we have
$$
\begin{array}{l}
k_{1}=(m-2)(n-2)(r-2), \\
k_{1}=2(m-2)(n-2)+2(n-2)(r-2)+2(r-2)(m-2), \\
k_{2}=4(m-2)+4(n-2)+4(r-2) .
\end{array}
$$

According to the given $k_{0}-k_{1}+k_{2}=1985$, we have
$$
\begin{array}{l}
(m-2)(n-2)(r-2)+4[(m-2)+(n-2)+(r-2)]-2[(m \\
-2)(n-2)+(n-2)(r-2)+(r-2)(m-2)]=1985 .
\end{array}
$$

Simplifying, we get
$$
(m-4)(n-4)(r-4)=1977 .
$$

This is an indeterminate equation about $m, n, r$, and we need to find the positive integer solutions that satisfy $r \geqslant n \geqslant m \geqslant 3$. Since $1977=1 \times 1 \times 1977=1 \times 3 \times 659$ and 659 is a prime number, we have
$$
\left\{\begin{array} { l } 
{ m - 4 = - 1 , } \\
{ n - 4 = - 1 , } \\
{ r - 4 = 1 9 7 7 , }
\end{array} \quad \left\{\begin{array} { l } 
{ m - 4 = 1 , } \\
{ n - 4 = 1 , } \\
{ r - 4 = 1 9 7 7 , }
\end{array} \quad \left\{\begin{array}{l}
m-4=1, \\
n-4=3, \\
r-4=659 .
\end{array}\right.\right.\right.
$$

This gives us 3 sets of solutions $(3,3,1981),(5,5,1981)$, and $(5,7,663)$. Summarizing, there are 5 sets of solutions: $(1,3,1987),(1,7,399),(3,3,1981),(5,5,1981)$, and $(5,7,663)$.