= Solution
Now the doublet <invariant tensor> would satisfy $D_1(Q)^T L D_2(Q)=L$. Its four entries acquire factors $\alpha^3$, $\alpha^{-1}$, $\alpha$, and $\alpha^{-3}$, respectively. None equals $1$, so $L=0$. Equivalently, $R_2\otimes R_3=R_2\oplus R_3$ has no <trivial representation>. The mixed doublet and third-family terms are again forbidden because their <tensor product of group representations> contain no <trivial representation>. Only the product of the two $R_1$ fields survives:
$$
\boxed{\lambda^{\mathrm{(ii)}}=
\begin{pmatrix}0&0&0\\0&0&0\\0&0&y_3\end{pmatrix},\qquad y_3\in\mathbb C.}
$$
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