Solution: Recall that \begin{align*} \phi _j(e_i) = \delta _{ij} = \begin{cases} 1, &\text{ if } i = j ;\\ 0, &\text{ if } \text{ otherwise}. \end{cases} \end{align*} If $\mathbf{x} = \left( x_1, x_2, \dots, x_5 \right) $, then $\phi _j(\mathbf{x} ) = x_j$. If $\phi \in \operatorname{span}\left\{ \phi _3,\phi _4,\phi _5 \right\} $, then we have \[ \phi = c_3\phi _3 + c_4\phi _4 + c_4\phi _5, ~~~c_3,c_4,c_5 \in \mathbb{R} . \]

Also, for any $\mathbf{x} = \left( x_1, x_2, 0,0,0 \right) \in U$ we have $\phi (\mathbf{x} ) = 0$. This implies, $\phi \in \mathrm{ann} (U)$ and hence $\operatorname{span}\left\{ \phi _3, \phi _4, \phi _5 \right\} \subseteq \mathrm{ann} (U) $. On the other hand, let $\phi \in \mathrm{ann} (U)$. Since $\phi \in \mathrm{ann} \subseteq \left( \mathbb{R} ^5 \right) ^*$, there exists constants $a_1,a_2,\dots,a_5$ such that \[ \phi = \sum_{i=1}^{5} a_{i} \phi _i. \] We need to show that $\phi \in \operatorname{span}\left\{ \phi _3, \phi _4, \phi _5 \right\} $, that is, $a_1= a_2 = 0$. Observe that $e_1, e_2 \in U $, so \begin{align*} \phi (e_1) = 0 \implies \sum_{i=1}^{5} \phi _i \left( e_1 \right) = a_1 \\ \phi (e_2) = 0 \implies \sum_{i=1}^{5} \phi _i \left( e_2 \right) = a_2. \end{align*} Therefore, $a_1 = a_2 = 0$ and hence $\phi \in \operatorname{span}\left\{ \phi _3, \phi _4, \phi _5 \right\} $. Thus, $\mathrm{ann} (U) \subseteq \operatorname{span}\left\{ \phi _3, \phi _4, \phi_5 \right\} $.