What is an example of a three-dimensional smooth distribution $D$ of $\mathbb{R}^4$ with this property:
Not only $D$ is not integrable but also there is no a two-dimensional foliation $F$ of $\mathbb{R}^4$ such that for every $x\in \mathbb{R}^4$, the tangent space to each leaf at $x$ is contained in $D_x$.
Now that this other question as been answered, we can answer this original question: The answer is 'yes' there exists a non-integrable tangent $3$-plane field on $\mathbb{R}^4$ that has no subordinate co-dimension $2$-foliation.
It suffices to take an open $p$-neighborhood $U$ diffeomorphic to $\mathbb{R}^4$ of a point $p$ that lies on the line $L$ described in my answer to that question. Then choosing a diffeomorphism $\phi:\mathbb{R}^4\to U$, set $\beta = \phi^*\alpha$. Then $\ker\beta$ will be a non-integrable tangent $3$-plane field on $\mathbb{R}^4$ that has no subordinate co-dimension $2$-foliation, because there is no such foliation on a neighborhood of $\phi^{-1}(p)$.
Of course, this says nothing about the following more precise question: Let $\beta$ be a $1$-form on $\mathbb{R}^4$ such that $\beta\wedge\mathrm{d}\beta$ is nowhere vanishing. Must there exist a co-dimension $2$ foliation of $\mathbb{R}^4$ whose leaves are all $\beta$-null?
I believe the standard example $dz - ydx = 0$ does the trick.
This is the normal form for what many people call a ``quasi-contact distribution'-the even dimensional analogue of a contact form. Suppose, by way of contradiction, that $F$ is one of your foliations and $L$
the leaf of $F$ through a point $x$. Then the pull back
of the above oneform to $L$ is zero.
But $d$ of the one form is $dx \wedge dy$, which, restricted to the distribution, is a rank 2 form, and so cannot have $T_x L = F_x$ in its kernel.
Contradiction.
