First,
there's only one way the 3 in the bottom left can be adjacent to three shaded cells.
And what about the 3 above it? It can't go up by two to place its other two shaded cells, because that would block off a square on the left wall. So it must go right two: this places two more shaded cells.
Now, since every tromino must have exactly one clue in it,
the unshaded cell at the bottom must be part of the 1 on the right.
And likewise, the cell above it can't be part of the 3, or it wouldn't be able to touch 3 gray cells -- so it must be part of the ? clue instead.
And some more reachability logic gets us here:
Next we can look at a different area of the puzzle:
The 3 in the top left has only one way to satisfy it without putting two shaded cells by the 1 clue.
And finally, one argument finishes it off:
There are 17 clues; this means there will be 13 shaded cells in all. We have 11 of them, so the 3 on the right must touch one of our already-placed cells. This can only be the one in the second row.
And then, we can place one shaded cell with the top 3; the bottom 3 only has one place its other shaded cell can go, and the puzzle is solved!
