Looking at r3c3, we realise the 3's targeting it much be in boxes 2, 4 & 5, and there are only two possible configurations for them:
OPTION 1: a 3 in r4c1, r5c4 & r2c5 or OPTION 2: a 3 in r5c2, r4c5 & r1c4
Either way, that means the green diamond in r2c2 is targeted by precisely 2 3's, which lets us place a "2" in it.
Looking at the r6c7, we realise the 5's targeting it can only be: Box 2: r3c5 Box 3: r2c8 One of: Box 4 (r5c3) or Box 7 (r7c3) and One pair of: Box 8 (r9c5) & Box 5 (r4c4) or Box 8 (r8c4) & Box 9 (r9c9)
Therefore, since we need 5 of those, we can straight away put a "5" into r2c8
Looking at r2c3 we can whittle down the possible values which would fit in the blue diamond.
It can't contain 1, 2, 3 or 5 (by sudoku). It can't contain 6, 7, 8 or 9 (because there are not that many boxes within distance 5 of it that could contain a 5 targeting the cell)
Therefore it must be a "4". And to have 4 5s at the right distance, there must be 5s in the cells r7c3 and r4c6
Indeed, we realise (belatedly) that the "5" in r3c5 eliminates the possibility of 5 in r9c5, and the 5 in r6c7 knocks out r6c2, which also lets us place all the remaining 5s: r1c2, r5c1, r9c9 and r8c4
Now we've placed the 5s, we can look at r5c2 and realise that there are already more than three cells targeting it: (the 3 at r3c3 and the 5s at r3c5, r4c6 and r8c4) which means it can't itself be a three. This lets us place 3s at r4c1, r5c4 and r2c5.
Adding in pencil marks in the remaining empty diamonds (just eliminating numbers they share a box, row or column with), gives us:
To narrow them down further, it would be nice to know where all the 7s and remaining 3s are (or as many as we can locate at this stage). The easiest way to do this is to put a 37 pair in every empty box not ruled out by standard sudoku:
And then spend two minutes eliminating every "3" that is 3 distance away from a yellow or blue cell, and every "7" that is 7 distance away from a red, green or blue cell:
The looks a lot more manageable and we notice that r1c7 contains a red diamond, which only counts 3s and 5s targeting it. We know all the 5s, and any unknown 3s are more than 3 distance from it, so we can work out that the complete list is: "5" at r1c2, "5" at r6c7 and "3" at r2c5. That's three cells targeting it, therefore the red diamond contains the digit 3.
Looking at the 7s in box 9, we spot that 2 of them are a distance of 7 away from the green 2 at r5c5 (which we should have noticed earlier), making r8c8 a 7, which starts a chain of eliminations:
If we look at the yellow 2 in r1c9, we see there are now only 2 cells left that are 7 distance away from it, so therefore both those cells (r3c4 and r4c5) must be "7"s.
Looking at r1c5 we don't know which colour it is, but we've identified all the 3s, 5s and 7s in range of it so we can see how high it could be even if they all counted:
3s - none target it 5s - none target it 7s - the only possible place left for a 7 to target it is r6c3, therefore it must a yellow "1" and r6c3 must indeed be a "7" (which forces r1c1 to be a "7" too). And that's now all the 7s placed:
Which means we can say for certain that r9c3 is 2 (it is only targeted by the 7s at r3c4 and r4c5), which lets us fill in the rest of the 2s via normal sudoku:
Looking at r5c2, the only uncertain 3, 5 or 7 within range of it is the potential 3 at r8c2. We are not told what colour it is, but we list the types of cell targeting at and so work that out:
3s - 1 (or possibly 2) target it 5s - 3 target it 7s - 1 targets it
So it is grey, and either a 5 or a 6.
But it can't be a five (there's already one in r5c1), so it must be a 6, which also implies that r8c2 is a "3". Following that through the chain of 3s we now have them all, and can also fill in the values for r7c8 and r7c9:
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