My solution path. Note that cells with a black dot in them are confirmed as unshaded. An L-mino is a corner-shaped tromino, and an I-mino is a straight tromino
There is only one way to fit 3 dominoes around the corner 3 and have them not touch. If the 3 is an L-mino then only 2 can touch, and if it is an I-mino going down then the dominoes can't fit due to the ? in R3C2. Therefore we have some shaded cells and some unshaded cells placed quickly.
For the dot in R2C2 to be part of a tromino, it must be part of an I-mino going down with the ? in R3C2. This also forces the dot in R4C1 to be a down I-mino, and it will join with the ? in R6C1.
No tromino can reach R7C1, so it must be shaded. This places another domino and some more dots.
Since shaded cells can't be isolated, R5C2 must have a dot. Only the ? in R5C3 can reach R5C2 and R6C2, so that must be an L-tromino
The only way that R4C3 can be part of a tromino is if it connects to R4C5 in a horizontal I-mino.
No tromino can reach R5C4, so it must be shaded. If its domino is horizontal then there is no way to satisfy the bottom-row 3. Therefore there is a domino going down, which allows the bottom-row 3 to have its tromino placed. The standard new dots along the edges of dominoes can also be placed.
There is now only one way to place the third domino for the bottom-row 3.
The dots at R6C5 and R5C5, by distance, must be part of the R6C6 ?'s tromino. This tromino can therefore be placed. Now the dot in R7C7 must extend up. The only way for it to be part of a tromino is for it to be a vertical I-mino with R5C7.
An intermediate global deduction:
There are three ?s left to make trominoes, and 13 cells not part of a tromino or domino. Therefore there must be 2 more dominoes
If R3C5 is shaded, the only way to place two dominoes leaves only 5 cells to split between R2C4's ? and R2C6's ?, so therefore R3C5 is unshaded. If R3C7 is shaded, then the only way to place two dominoes leaves R4C7 with only one cell. Therefore R3C7 is unshaded.
It is now clear how the 2 remaining dominoes must be placed. The one in C6 must extend to R5C6 in order to not orphan it as an unshaded cell.
Step 11 (last step):
The ? in R4C7 must use the dot in R3C6 in order to be a tromino; that places its tromino. The ? in R2C6 must therefore use the dot in R2C5 to be a tromino. That completes all the borders, and therefore the puzzle.