How to prove the NP-completeness of the ``Exact-3D-Matching`` problem by reducing the ``3-Partition`` problem to it?
I think this reduction should work:

Given a 3-PARTITION instance: $n$ integers $a_1,\ldots,a_{3n}$ and a target sum $B$; 
mark each integer with an unique identifier $\hat{a_i}$; then build a collection of 3-elements subsets $C_j = \{ \hat{a}_{i_1}, \hat{a}_{i_2}, \hat{a}_{i_3} \}$ such that $a_{i_1}+ a_{i_2}+ a_{i_3} = B$ and $i_1 \neq i_2 \neq i_3$ (the collection has $O(n^3)$ elements).

The original 3-PARTITION problem has a solution if and only if there exist an EXACT-3D-MATCHING $C_{j_1}, \ldots, C_{j_n}$ that covers $S = \hat{a}_{1}, \ldots,  \hat{a}_{3n}$.

For example given $A = \{5,4,3,3,3,2,2,1,1\}, B = 8$ we build:
$S = \{5a,4a,3a,3b,3c,2a,2b,1a,1b\}$ and the collection of 3-elements subsets:

C1  = { 5a, 2a, 1a }
C2  = { 5a, 2a, 1b }
C3  = { 5a, 2b, 1a }
C4  = { 5a, 2b, 1b }
C5  = { 4a, 3a, 1a }
C6  = { 4a, 3a, 1b }
C7  = { 4a, 3b, 1a }
C8  = { 4a, 3b, 1b }
C9  = { 4a, 3c, 1a }
C10 = { 4a, 3c, 1b }
C11 = { 4a, 2a, 2b }
C12 = { 3a, 3b, 2a }
C13 = { 3a, 3b, 2b }
C14 = { 3a, 3c, 2a }
C15 = { 3a, 3c, 2b }
C16 = { 3b, 3c, 2a }
C17 = { 3b, 3c, 2b }


An Exact-3D-MATCHING is given by:

C1  = { 5a, 2a, 1a }
C6  = { 4a, 3a, 1b }
C16 = { 3b, 3c, 2b }


that also uniquely identifies a valid solution for the original 3-PARTITION problem.