= Solution
The eight <Gell-Mann matrices> generate $G=SU(3)$, and the three superfields form one fundamental triplet. A renormalizable <superpotential> may have degree at most three. There is no invariant vector or symmetric rank-two tensor in the fundamental. The only apparent cubic singlet is
$$
\epsilon^{ijk}\Phi_i\Phi_j\Phi_k,
$$
but chiral superfields are Grassmann-even and commute, so this contraction of a symmetric product with $\epsilon^{ijk}$ vanishes identically. Consequently the most general invariant superpotential as the question is written is only a constant,
$$
W=W_0,
$$
which has no effect in global supersymmetry. In particular, $W=\lambda\Phi_1\Phi_2\Phi_3$ is not invariant under a general $SU(3)$ rotation.
Solved by gpt-5.6-sol high.
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