Numerical 3-dimensional matching

Take ${\displaystyle X=\{3,4,4\}}$, ${\displaystyle Y=\{1,4,6\}}$ and ${\displaystyle Z=\{1,2,5\}}$, and ${\displaystyle b=10}$. This instance has a solution, namely ${\displaystyle \{(3,6,1),(4,4,2),(4,1,5)\}}$. Note that each triple sums to ${\displaystyle b=10}$. The set ${\displaystyle \{(3,6,1),(3,4,2),(4,1,5)\}}$ is not a solution for several reasons: not every number is used (a ${\displaystyle 4\in X}$ is missing), a number is used too often (the ${\displaystyle 3\in X}$) and not every triple sums to ${\displaystyle b}$ (since ${\displaystyle 3+4+2=9\neq b=10}$). However, there is at least one solution to this problem, which is the property we are interested in with decision problems. If we would take ${\displaystyle b=11}$ for the same ${\displaystyle X}$, ${\displaystyle Y}$ and ${\displaystyle Z}$, this problem would have no solution (all numbers sum to ${\displaystyle 30}$, which is not equal to ${\displaystyle k\cdot b=33}$ in this case).

Every instance of the Numerical 3-dimensional matching problem is an instance of both the 3-partition problem, and the 3-dimensional matching problem.

NP-completeness of the 3-partition problem is stated by Garey and Johnson in "Computers and Intractability; A Guide to the Theory of NP-Completeness", which references to this problem with the code [SP16].[1] It is done by a reduction from 3-dimensional matching via 4-partition. To prove NP-completeness of the numerical 3-dimensional matching, the proof is similar, but a reduction from 3-dimensional matching via the numerical 4-dimensional matching problem should be used.