= Solution
Let $\nu=-L_3$, whose <Dynkin labels> are $(0,1)$. Since $\lambda-\nu=L_1-L_3$ is the sum of the two <simple roots>, its boundary distances in the <dominant weight multiplicity formula for sl3> are $r+1,s+1$. Consequently
$$
\dim R(a,b)_\nu=1+\min(a,b,r+1,s+1).
$$
For a module in $\mathcal C_1$, this multiplicity is one in the two families having $a=0$ or $b=0$, and two for $R(1,2)$ and $R(3,1)$. For a module in $\mathcal C_2$, it is two in the families having $a=1$ or $b=1$, and three for $R(2,3)$ and $R(4,2)$. In each case, increasing $r,s$ by one increases the minimum only if neither $a$ nor $b$ already limits it.
To obtain total multiplicity five at $\nu$, the two summands must therefore contribute two and three. \b[The only surviving pairs, up to interchange, are]
$$
\boxed{\begin{aligned}
&\{R(1,2),R(2,3)\},&&\{R(1,2),R(4,2)\},\\
&\{R(3,1),R(2,3)\},&&\{R(3,1),R(4,2)\}.
\end{aligned}}
$$
Their multiplicities at $\lambda$ remain one and two, so all four pairs satisfy both conditions.
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