A strongly non-integrable distribution 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 $D_x$.

 A: 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?
A: 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.  
