The solution to the puzzle is as follows:
Steps to solving:
Firstly, notice that the 3 cannot be a straight line otherwise we would have to have a wall of 3 consecutive shaded cells.
Filling in some cells adjacent to trimonoes and the 3 gives us this:
To avoid a shaded L-tetromino, R10C5 must be shaded. If that is part of a top-left corner L-trimono, the 7 region is too large. So R10C6 is also shaded.
Now, consider what happens if R9C5 is shaded. A few more cells must be shaded to block off the 7 as shown:
We get some more cells from the central 3:
If R7C7 was shaded, then the right 3 would become an X-pentomino. So R9C7 is shaded instead. However, this results in an I-trimono being shaded to prevent the 4 from becoming an F-pentomino:
Thus, R9C5 is not shaded, and R9C6 therefore has to be. This gives another shaded cell to block off the 7. R9C8 also has to be shaded, otherwise the 4 becomes an L-tetromino which would result in an adjacent shaded I-trimono.
If R8C8-9 were shaded, the 4 would either have only 1 cell or at least 11 cells, depending on whether there is a trimono in the corner.
If R8C7-8 were shaded, the 4 would either have 3 or 9 cells, depending on whether there is a trimono in the corner.
Thus, R10C8-9 are shaded. (Also, R6C6 is shaded to prevent the 3 from becoming a V-pentomino.) Then if R9C10 were shaded the resultant trimono would make a one-cell 4 region, so it must be unshaded.
If the 4 were an S-tetromino, it would join with the unshaded R8C8 to become a W-pentomino. So it is instead a T-tetromino, as shown:
Several trimonoes are now forced, to block off the 3 regions:
Now, we need to block off the 4 so that it doesn't become a 7-cell snake:
If R3C5 were shaded, to avoid the 5 from becoming a 7-cell tree we need to shade R2C6 which would then form an at least 4-cell shaded block.
So R3C5 is unshaded.
If R2C5 were unshaded, note that since either R7C1 or R7C2 are shaded the 5 must escape down or left, and it would thus connect to form a 7-cell tree.
So R2C5 is shaded. Also, R3C3 is shaded to complete a trimono.
If R1C4-5 are shaded, then to stop the 5 from becoming a P-hexomino, R3C6 is also shaded. But then the resultant trimono blocks off the 5 to a region of at most 4 cells.
If R1C5 and R2C6 are shaded, then the trimono to the right of 5 needs R1C7 shaded too, blocking the 5 into a monomino.
Thus, R2-3C6 are shaded, forcing R1C7 shaded in the trimono to the 5's right as well.
Finally, the 5 must form an N-pentomino, and the last trimono is forced: