Is there a $2$- dimensional foliation of $\mathbb{R}^4\setminus \{0\}$ whose tangent space is contained in $\ker \alpha$ where $\alpha$ is the following non integrable $1$-form?


$$\alpha=(x^2+y^2)dx+(xz+yw)dy+(xz+yw)dz+(z^2+w^2)dw$$


In the matrix language, the $1$-form $\alpha$ can be counted as $\alpha=trace(AA^{tr}dA)$ where $$A=\begin{pmatrix}x&y\\ z&w \end{pmatrix}$$



Is there a $2$ dimensional foliation tangent to this particular $3$ dimensional distribution?
There is no co-dimension 2 foliation of the desired kind in a neighborhood of any point on the line $L\subset\mathbb{R}^4$ defined by $x-w=y=z=0$. Thus, if one wants to find such a foliation, it will be necessary to remove this entire line from the domain.


Here is the reason: Consider the $3$-form $\Upsilon = \alpha \wedge \mathrm{d}\alpha$. It vanishes along the line $L$. When one expands it in the $wxyz$-coordinates, the coefficients of this $3$-form are cubic polynomials in $wxyz$.


Now, suppose that there were a codimension $2$ foliation of the desired kind in a neighborhood of a point $p\in\mathbb{R}^4$ and let it be defined as the simultaneous level sets of two functions $f$ and $g$ defined in a $p$-neighborhood $U\subset\mathbb{R}^4$ and independent there: $\mathrm{d}f\wedge\mathrm{d}g$ is nonvanishing on $U$. The condition that $\alpha$ vanish when pulled back to the leaves of this foliation is that $\alpha\wedge\mathrm{d}f\wedge\mathrm{d}g$ vanish identically. This also implies that, when $\alpha$ is pulled back to a level set of $\mathrm{d}f$, it must become integrable, i.e., $\Upsilon = \alpha \wedge \mathrm{d}\alpha$ must vanish when pulled back to a level set of $f$, i.e., $\alpha\wedge\mathrm{d}\alpha\wedge\mathrm{d}f$ vanishes identically on $U$. Similarly, $\alpha\wedge\mathrm{d}\alpha\wedge\mathrm{d}g$ must vanish identically on $U$. 


Now, using a Taylor series expansion about $p$, it is easy to show that, when $p$ lies on the line $L$, the equation $\alpha \wedge \mathrm{d}\alpha\wedge \mathrm{d}f = 0$ implies that, at $p$, $\mathrm{d}f$ must be a multiple of $\mathrm{d}(x{+}w)$. Similarly, at $p$, $\mathrm{d}g$ must be a multiple of $\mathrm{d}(x{+}w)$. Thus, $\mathrm{d}f\wedge \mathrm{d}g$ must vanish at $p$, contrary to the way $f$ and $g$ were chosen.


Thus, the claim is established.


*Remark*: When $p$ is the origin, the Taylor series expansion shows that $\mathrm{d}f$ must actually vanish at the origin, so the exclusion is even stronger there.


