Two conjugate SU(3) triplets with a triplet ($\overline{3}\otimes \overline{3}\otimes 3=2\overline{3}\oplus 6\oplus V_{(1,2)}$)

Split the two dual factors into antisymmetric and symmetric tensors. The invariant Levi-Civita symbol identifies the antisymmetric pair with $3$, giving $6\oplus \overline{3}$ after multiplying by the remaining triplet. On the symmetric part, contraction $T^a{}_{ba}$ maps onto $\overline{3}$; the equivariant splitting is $T^a{}_{bc}=({\delta }_b^a{v}_c+{\delta }_c^a{v}_b)/4$. Its trace-free kernel has dimension $18-3=15$ and highest weight with Dynkin labels $(1,2)$. The Weyl dimension formula gives dimension $15$, so this is exactly the irreducible kernel. It is conjugate to the other fifteen-dimensional module with labels $(2,1)$; dimension alone does not distinguish the two.