I've tried to write this walkthrough in a Paper Solvable way; there may be a few places where I've been a bit lazy and not specifically restated something which obtained from an earlier step. I've deliberately avoided using the term Hidden Single; maybe other solvers use it for Paper Solvable solving but it seems to me to imply elimination solving.
There are three cages on each diagonal, each containing the same three numbers. There is also an Old Lace cage centred on R5C5.
1. 24(3) cages must be {789}, 20(3) cages must be {569} (only combination which can overlap with 24(3) in N5 without clashing with it) -> R5C5 = 9
2. The four outer Old Lace cells must contain 1,2,3,4; the four zero cells in N5 must contain 1,2,3,4
2a. The two outer Old Lace cells in R5 and the two zero cells in R5N5 must contain {1234}, 11(4) cage at R4C7 only contains {1235} -> R5C8 = 5
2b. The two outer Old Lace cells in C5 and the two zero cells in C5N5 must contain {1234}, 11(4) cage at R4C4 only contains {1235} -> R2C5 = 5
3. 12(4) cage at R5C2 must contain four of {123456}, 5 already placed for R5 with {1234} required for two outer Old Lace cells in R5 and the two zero cells in R5N5 -> R5C2 = 6
4. The four outer cells of the Old Lace must be {1234} but cannot have 4 in the 11(4) cages or in the 12(4) cage (because 6 placed so other cells must contain 1,2,3) -> R7C5 = 4 (only remaining place for 4 in Old Lace)
5. 13(4) cage at R7C5 containing 4 must contain two of 1,2,3 and one of 5,6 but 1,2,3 required for remaining Old Lace cell in C5 and two zero cells in C5N5 -> R8C5 = 6
6. 18(4) cage at R1C6 can only be {1467} (cannot contain more than one of 1,2,3 because 11(4) cage at R4C7 requires two of 1,2,3 for C7; cannot place any of 1,2,3 in C6 because they are required for 11(4) cage at R2C4, cannot contain 5 because R2C5 = 5 and 5,6,9 in N3 required for 20(3) cage at R1C9) -> R12C6 contain two of {467}, 6 locked for C6 -> R4C6 = 5, R6C4 = 6, R12C7 contain two of {147}, 1 locked for C7 -> R4C8 = 1 (only remaining place for 1 in 11(4) cage at R4C7
7. R4C9 = 6 (only remaining place for 6 in R4, because 11(4) cage at R4C7 cannot contain 6)
8. R6C1 = 5 (only remaining place for 5 in R6 because 12(4) cage at R5C2 cannot contain both of 5,6)
9. 18(4) cage at R3C1 must contain one of 1,2,3 in R3C12 or cage total would be more than 18 (cannot contain two of 1,2,3 because R3C56 require two of 1,2,3 for 11(4) cage at R2C3, cannot contain 2 or 3 in R4C12 because 12(4) cage at R5C2 with 6 requires all of 1,2,3 for N4) must contain one of 7,8 but not 9 (because 1,4,5,9 is greater than 18) -> R4C3 = 9 (only remaining place for 9 in R4)
10. 18(4) cage at R3C1 must contain one of 1,2,3 in R3C12 (step 9), R3C45 require two of 1,2,3 for 11(4) cage at R2C4, R12C7 contain two of {147} (step 6) -> 2,3 for N3 can only be in R1C8 + R2C9 -> 8 in N3 must be in R3C89, locked for R3 -> R3C3 = 7 (because 24(3) cage at R1C1 requires one of 7,8,9 in R3C3)
11. 2,3 in N3 can only be in R1C8 + R2C9 (step 10) -> 1,7 on N3 can only be in R12C7 (because 1,2,3 required in 18(4) cage at R3C1 and R3C45 for R3, step 10) -> R12C6 must contain 4,6, locked for C6 -> R5C6 = 3 (because 1,2 required in R78C6 for 13(4) cage at R7C5), R45C7 must be [32] for 11(3) cage at R4C7, R4C5 must be 2 for N5, then R6C5 = 1, R5C4 = 4, R5C3 = 1 (only remaining place for 1 in 11(4) cage at R5C2), R3C5 = 3 (required for 11(4) cage at R2C4 and 1,2,5 already placed for C5)
11a. 7 in R2 can only be in R2C7(because 1,7 in R12C7 for N3 and 18(4) cage at R1C6, see earlier in step 11) -> R12C7 = [17]
12. R9C2 = 1 (only remaining place for 1 in R9, R9C1 requires one of {69} for 20(3) cage at R7C3, R78C6 require {12} for 11(4) cage at R7C5, R9C9 requires one of {789} for 24(3) cage at R7C7)
13. R9C5 requires one of {78} for C5, R9C6 requires one of {789} for C6, R9C9 requires one of {789} for 24(3) cage at R7C7, 20(3) cage at R7C3 contains {569} -> R9C1 = 6, R7C3 = 5, R8C2 = 9
13a. 24(3) cage at R1C1 = {789} -> R2C2 = 8, R1C1 = 9
14. R3C6 = 9 (only remaining place in N2, 18(4) cage at R1C6 cannot contain 9 because R12C7 = [17], 11(4) cage at R2C4 cannot contain 9)
15. 20(3) cage at R1C9 contains {569} -> R1C9 = 5, R3C7 = 6, R2C8 = 9
16. R9C5 requires one of {78} for C5, R9C6 requires one of {78} for C6, R9C9 requires one of {789} for 24(3) cage at R7C7 -> R9C9 = 9, R7C7 = 8, R8C8 = 7
16a. R6C7 = 9 (only remaining place in C7)
17. R7C8 = 6 (only remaining place for 6 in C8)
17a. 18(4) cage at R6C8 = {1467} (only remaining combination, cannot be {1368/2367/3456} because 1,2,3,5 not available in R6C89) -> R6C8 = 4, R67C9 = [71], R6C6 = 8, R4C4 = 7, R9C56 = [87]
18. 13(4) cage at R7C5 requires {12} in R78C6 -> R78C6 = [21]
19. R89C4 can only contain {35} = 8 -> R89C3 = 12 = {48} (only remaining combination) -> R89C3 = [84]
and the rest is straightforward, the equivalent of naked singles in elimination solving.
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A true Paper Solvable; after each placement there seemed to only one way forward for the next step.