= Solution
The short-root <Weyl reflection> $s_\alpha$ interchanges $\omega_1$ and $\omega_2$, because $\alpha=\omega_2-\omega_1$ and $\alpha^\vee=\delta_2^\vee-\delta_1^\vee$. Hence
$$
s_\alpha\lambda=b\omega_1+a\omega_2.
$$
The <Weyl group> orbit of the highest weight occurs in $U$ with a one-dimensional <extremal weight space>. Let $v$ be a nonzero vector of weight $s_\alpha\lambda$. The same reflection exchanges the long simple roots: $s_\alpha\delta_1=\delta_2$ and $s_\alpha\delta_2=\delta_1$. If an $A_2$ raising operator did not annihilate $v$, then $s_\alpha\lambda+\delta_i$ would be a weight of $U$; applying $s_\alpha$ would make $\lambda+\delta_j$ a weight, contradicting the fact that $\lambda$ is the <highest weight>.
Thus $v$ is an $A_2$ <highest-weight vector> of weight $b\omega_1+a\omega_2$. The <Complete reducibility of semisimple Lie algebra representations> then supplies the corresponding irreducible summand, so
$$
\boxed{V(b\omega_1+a\omega_2)\subseteq U\!\downarrow_{\mathfrak{sl}_3}.}
$$
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