There are $27$ different possibilities to enter into the lock. Let's say two of them are connected if they have two common digits (in the same positions); so every possibility is connected to exactly $6$ others. We can show them on a diagram like this:
The single-digit possibilities are shown in red, and the double-digit possibilities in green. Each single-digit possibility is connected to six double-digit ones around it. Each double-digit possibility is connected to one single-digit, one double-digit in red (same digit appearing twice), two double-digits in green (same two digits appearing), and two triple-digit possibilities (not shown on this diagram).
We can also see here that two triple-digit possibilities which are cycled versions of each other must be connected to two disjoint collections of double-digit possibilities:
So three cycled triple-digit possibilities are enough to connect to ALL the double-digit possibilities, but the single-digit possibilities are still separate. That means we could unlock the safe with
six tries: e.g. $012,201,120,000,111,222$.
But we can do even better by using the "middle" type of possibility: not triple and single, but use the double-digit ones that connect to both. Any trio of double-digit possibilities (we can see they're arranged into trios) will, between them, connect to ALL the triple-digit possibilities, and also to all the other double-digit possibilities which either use the same two digits or use the same digit twice, as well as of course one single-digit possibility. For example,
trying $100,010,001$ will cover all triple-digit possibilities and also $110,101,011$ and $002,020,200$ and $000$. Now the remaining double-digit possibilities are $112,121,211$ and $122,212,221$ and $022,202,220$ which are covered by $111$ and $222$.
So we can unlock the safe with
five tries: $100,010,001,111,222$.
Is it possible to do better?
No, see hexomino's answer. (I found this solution independently, but didn't get around to proving optimality before the other answer was posted. However, I still think it's worth having this answer for the pictorial approach which makes it seem natural.)
We know that every triple-digit possibility is connected to exactly one double-digit from each trio, and every double-digit possibility is connected to exactly two triple-digit possibilities which are not cycles of each other (and are therefore transpositions of each other). Each pair of transposed triple-digit possibilities has two different double-digit possibilities connected to the same pair (e.g. $012,102$ are both connected to both $112$ and $002$).
So we can manage to deduce the exact password with
8 tries: all the triple-digit possibilities and two of the single-digit ones. After trying all the triple-digit ones, we know that: if only one of them works, then that's the exact password; if exactly two of them work, then we have two double-digit possibilities for the password; if none of them work, then we have three single-digit possibilities. In either of the latter two cases, we can distinguish between remaining possibilities by trying two of the single-digit ones.