7.109 Given that the lock on a safe is composed of 3 dials, each with 8 different positions. Due to disrepair, the safe can now be opened as long as two out of the three dials are in the correct position. How many combinations must be tried at a minimum to ensure the safe can be opened?
For each knob, the 8 different positions are denoted as $1,2, \cdots, 8$, and each combination is denoted as $(i, j, k)$, where $1 \leqslant i, j, k \leqslant 8$. Below, using the dictionary order, we write down 16 triples $(i, j, k)$ for $1 \leqslant i, j, k \leqslant 4$ such that any two triples have at most one number in common. By adding 4 to each component of these triples, we obtain another 16 triples, making a total of 32 triples as follows:
$$
\begin{array}{l}
(1,1,1),(2,1,2),(3,1,3),(4,1,4), \\
(1,2,2),(2,2,1),(3,2,4),(4,2,3), \\
(1,3,3),(2,3,4),(3,3,1),(4,3,2), \\
(1,4,4),(2,4,3),(3,4,2),(4,4,1), \\
(5,5,5),(6,5,6),(7,5,7),(8,5,8), \\
(5,6,6),(6,6,5),(7,6,8),(8,6,7), \\
(5,7,7),(6,7,8),(7,7,5),(8,7,6), \\
(5,8,8),(6,8,7),(7,8,6),(8,8,5) .
\end{array}
$$

For any combination $(i, j, k)$ of the 3 knob positions, by symmetry, we can assume $1 \leqslant i, j \leqslant 4$. It is easy to see that $(i, j)$ must appear in the first two components of one of the first 16 triples, so testing these 32 combinations will definitely open the cabinet.
On the other hand, testing 31 times does not guarantee that the cabinet will be opened.
Let $K$ be the set of these 31 triples. If two triples have at least two components in common, we say one covers the other. Testing these 31 times to ensure the cabinet opens is equivalent to the 31 triples in $K$ covering all possible triples.

We consider each triple as an integer point in three-dimensional space, then all possible $8^{3}$ triples correspond to all integer points $(i, j, k)$, $1 \leqslant i, j, k \leqslant 8$ within and on the surface of a cube with side length 7. Clearly, each integer point covers 21 other integer points, which are the 7 points on each of the 3 lines parallel to the 3 coordinate axes passing through the point $(a, b, c)$.

We color the points in $K$ red, then the 31 red points are distributed on 8 planes: $z=1,2, \cdots, 8$. By the pigeonhole principle, there must be a plane with at most 3 red points. Without loss of generality, assume this plane is $z=1$ and the 3 red points are $(1,1,1),(2,2,1)$, and $(3,3,1)$ (other cases cover fewer points than this one). Thus, there are 25 points
$$
(i, j, k), 4 \leqslant i, j \leqslant 8, k=1
$$

not covered by these 3 red points. Therefore, these 25 points can only be covered by red points on other planes. Since one red point can only cover one of these 25 points, 25 red points are needed to cover them all. In addition to these 25 red points, $K$ has 6 more red points, so the red points $(a, b, c)$ satisfying $1 \leqslant a, b \leqslant 3$ are at most 6. Thus, among the 8 sets
$$
P_{k}=\{(i, j, k) \mid 1 \leqslant i, j \leqslant 3\}, k=1,2, \cdots, 8
$$

there must be one set without red points, and thus the 9 integer points in this set need 9 red points to cover, which is impossible.
In summary, at least 32 tests are required to ensure the cabinet is opened.