Finding the maximum number of subspaces of a vector space over finite field that satisfy these relations I have a question and I am stuck. I was wondering if anyone has a thought, before I start a brute-force search.

For $q$ a prime number and $n =6$, let $\mathbb {F}_{q}^{n}$ be an $n$-dimensional vector space over $\mathbb{F}_{q}$.
  Furthermore, let $ U_1, \dots, U_m$ be a family of $2$-dimensional subspaces of $\mathbb{F}_{q}^n$ such that $U_i \cap U_j = \{0\}$ and $\langle U_i, U_j \rangle \cap U_k = \{0\}$, for all $i, j, k \in \{1,\dots, n\}$, $i\neq j \neq k$. What is the biggest possible $m$?

Thanks in advance.
 A: Let $q$ be a prime power and let $\mathcal{U}$ be a collection of $2$-dimensional subspaces of $\Bbb{F}_q^6$ such that for any three distinct planes $U_1,U_2,U_3\in\mathcal{U}$ we have
$$U_1\cap U_2=0\qquad\text{ and }\qquad (U_1+U_2)\cap U_3=0.$$
Then $\dim(U_1+U_2)=4$ and $\dim(U_1+U_2+U_3)=6$. Pick $U_0\in\mathcal{U}$ and let $\mathcal{U}':=\mathcal{U}-\{U_0\}$. Then for every $U\in\mathcal{U}'$ the quotient $(U_0+U_1)/U_0$ is a $2$-dimensional subspace of $\Bbb{F}_q^6/U_0$.
Lemma: For $U_1,U_2\in\mathcal{U}'$ with $U_1\neq U_2$ we have
$$(U_0+U_1)/U_0\ \cap\ (U_0+U_2)/U_0=0.$$
Proof. It suffices to note that $(U_0+U_1)\cap(U_0+U_2)=U_0$, which follows from the fact that 
$$\dim((U_0+U_1)+(U_0+U_2))=\dim(U_0+U_1+U_2)=6.$$
Corollary: The set
$$X:=\Big\{(U_0+U)/U_0-\{0\}:\ U\in\mathcal{U}'\Big\},$$
is a collection of disjoint subsets of $\Bbb{F}_q^6/U_0-\{0\}$ with $|X|=|\mathcal{U}'|$.
As noted $\dim((U_0+U)/U_0)=2$ and clearly $\dim(\Bbb{F}_q^6/U_0)=4$ so it follows that
$$(q^2-1)|X|\leq q^4-1,$$
or equivalently $|\mathcal{U}'|\leq q^2+1$ and hence $|\mathcal{U}|\leq q^2+2$.

The comments by JyrkiLahtonen and azimut indicate constructions of collections $\mathcal{U}$ with $|\mathcal{U}|=q^2+1$ for odd $q$, and $|\mathcal{U}|=q^2+2$ for even $q$. So this leaves only the question:

Is there such a collection $\mathcal{U}$ with $|\mathcal{U}|=q^2+2$ when $q$ is odd?

If, for example, projective $3$-space over $\Bbb{F}_q$ can be partitioned into projective lines, then the answer is yes.

Old answer:
Let $\mathcal{U}$ be a set of planes in $\Bbb{F}_q^6$ such that for any three planes $U_1,U_2,U_3\in\mathcal{U}$ we have
$$U_i\cap U_j=\{0\}\qquad\text{ and }\qquad\langle U_i,U_j\rangle\cap U_k=\{0\}.$$
Then $\langle U_i,U_j,U_k\rangle=\Bbb{F}_q^6$, so no three planes in $\mathcal{U}$ are contained in a single $5$-dimensional subspace. Every pair $U_i,U_j\in\mathcal{U}$ spans a $4$-dimensional subspace, and there are $q+1$ different $5$-dimensional subspaces containing it. Then no other plane in $\mathcal{U}$ is contained in any of these $q+1$ subspaces. This holds for any pair of planes in $\mathcal{U}$. The total number of $5$-dimensional subspaces of $\Bbb{F}_q^6$ is
$$\frac{q^6-1}{q-1}=q^5+q^4+q^3+q^2+q+1,$$
so the number of pairs of planes in $\mathcal{U}$ is at most
$$\frac{q^5+q^4+q^3+q^2+q+1}{q+1}=q^4+q^2+1.$$
Let $m:=|\mathcal{U}|$ be the number of planes in $\mathcal{U}$. Then the above says that
$$\binom{m}{2}\leq q^4+q^2+1,$$
where $\tbinom{m}{2}=\tfrac{1}{2}m(m-1)$. The quadratic formula then tells us that
$$m\leq\frac{1}{2}+\frac{1}{2}\sqrt{1+8(q^4+q^2+1)}=\frac{1}{2}+\sqrt{2q^4+2q^2+\tfrac{9}{4}}.$$
This gives us an upper bound for the biggest possible $m$, which is close to the lower bound of $q^2+1\leq m$ given in the comments in the sense that
$$\frac{1}{2}+\sqrt{2q^4+2q^2+\tfrac{9}{4}}\leq\frac{1}{2}+\sqrt{2}(q^2+1).$$
