Solution: Note that a basis for $U$ can be given as \[ \mathcal{B} _U = \{ (1,0,1,1,2), (0,1,-1,1,0) \}. \] Since, we need to find subspaces $U_1,U_2,U_3$ such that $\mathbb{R} ^5 = U \oplus U_1 \oplus U_2$, each of the $U_i$ should have one dimensional. We will start with a choice of $U_1$ as follows. Take a vector $\mathbf{v}_1 \in \mathbb{R} ^5$ which is not in $U$, then $U_1 = \operatorname{span}\left\{ \mathbf{v}_1 \right\} $. Note that $\mathbf{v} _1 = (1,0,0,0,0) \notin U$ as if $\mathbf{v} _1 \in U$, then there exists $\alpha ,\beta \in \mathbb{R} $ such that \[ \left( \alpha , \beta , \alpha -\beta ,\alpha +\beta, 2\alpha \right) = (1,0,0,0,0) \] which does not have a solution.

A similar strategy will be follow to find the subspaces $U_2$ and $U_3$. We choose $\mathbf{v}_2 \in \mathbb{R} ^5$ such that $\mathbf{v} _2 \notin \operatorname{span}\left\{ \mathcal{B} _U \cup \{ \mathbf{v} _1 \} \right\}$. We choose $\mathbf{v} _2 = (0,1,0,0,0)$. Again note that $\mathbf{v} _2 \notin \operatorname{span}\left\{ \mathcal{B} _U \cup \{ \mathbf{v} _1 \} \right\} $. So, $U_2 = \operatorname{span}\left\{ \mathbf{v} _2 \right\} $. Similarly, we take $\mathbf{v} _3 = (0,0,1,0,0)$ and note that $\mathbf{v} _3 \notin \operatorname{span}\left\{ \mathcal{B}_U \cup \{ \mathbf{v} _1 \cup \{ \mathbf{v} _2 \} \} \right\} $.

What we did in this process is we extended the set $\mathcal{B} _U$ to a basis of $\mathbb{R} ^5$. Since we need at least $5$ vectors in the basis, so we assigned span of each of the remaining three vectors to the subspaces $U_i$. Another way to find $U_1,U_2$ and $U_3$ is to find the orthogonal complement of $U$, which will be a subspace and the dimension will be three, let us say $\left\{ \mathbf{v} _1, \mathbf{v} _2 \mathbf{v}_3 \right\} $ is a basis of $U^\perp$ which is the orthogonal complement of $U$. Now we take \[ U_i = \operatorname{span}\left\{ \mathbf{v} _i \right\} ,~ i=1,2,3. \]