For starters, we can make the following deductions.
The 1 x 3 group in row 8, columns 6-8 can deduced immediately due to the '8' in R6C6. As a result, we can also place a '5' in R9C8 and an '8' in R1C8, due to the fact that they cannot appear in the 1 x 3 groups in column 8. 7 must belong to the 1 x 3 group in row 8, columns 3-5, and it must be in R8C4 due to '7' already in column 5. The '147' group must the 2nd 1 x 3 group in column 8, because the '4's eliminates any possibility of it being in the first. This allows for the placement for '7' in R6C8.
Next, we can fill in cells where there is only one possibility (8 in R4C9, 5 in R7C1, 7 in R9C7). 6 can be placed in R8C9, because there isn't a '369' group in row 8 and 3 and 9 are already in the bottom right square. Now, note that there can't be a '369' group horizontally in row 2. This means the square in R2C9 must be 3,6 or 9, but since 3 and 6 are already placed, it must be a 9. In R2C1, it must be a '3' since 6 is already there. The knock-on deduction gets us to:
At this point, I got stuck. However, I noticed that the '147' and '258' groups must appear somewhere in row 2. Depending on which one is the first group horizontally in that row 2, the places for the 1 x 3 cells in column 2 can be deduced as well. So, I decided to try that '147' group must appear in the 2nd '1 x 3' cell in row 2. (If this results in a contradiction, then we will know '258' should appear in that group instead). Making this try gives us the following grid based on process of elimination.:
Next, using the grid before, we can use process of elimination to place some more 1,4,7 and in the process also solve for the 1st 1 x 3 group in row 2.
Following that, we can note that '5' can only go in R3C5 and '8' can only go in R3C4. Additionally, the cell in R1C2 must be a 2,5,8, but since 5,8 are already filled, it must be a 2. This and other deductions get us to:
Next, note that in row 8, 3 must appear in R8C2 since there is already a '3' in Column 1. This allows us to place 9 in R8C1 and also solve for the last 1 x 3 group in column 2. This also allows us to place a '9' in R9C5 and some other knock-on deductions. So far, no contradictions!:
Using some more process of elimination logic, we can place a '8' in R5C1 and '2' in R7C6. Using the knock-on deductions from there:
Next, we can notice that in column 5, only 2 and 6 are left. This means '3' must appear in R5C6. This deduction allows us to completely finish writing in all the '3's. Next, in row 5, there is also a 2 and 6 missing, but since 6 already appears in column 3, 2 must appear in R5C3. This allows us to place a 6 in R5C7. Based on the chain deductions that follow from there, we can fill in all the remaining digits and there were no contradictions. So this is the final solution:
I decided that I show the other attempt as well to show that one leads to a contradiction and the above is the only solution to the problem.
Placing '258' in the second 1 x 3 group in row 2 will eventually lead to the following contradiction:
As can be seen, the only remaining numbers that can be placed in R3C4 are '1' and '7' but both of them already appear in the same column. Thus, the second 1 x 3 group in row 2 must be '147' as shown before.