Verifying the integrability of the distribution $\omega(x,y,z)=\operatorname{ker}(x\,dx + y\,dy + z\,dz)$ on $\Bbb R^3$ 
Consider the linear functional $\omega \in (\mathbb{R}^3)^{*}$, gived by $\omega(x,y,z)=x\cdot dx + y\cdot dy + z\cdot dz$ (The base $\{ dx,dy,dz \}$ for $(\mathbb{R}^3)^{*}$is dual in the canonical base of $\mathbb{R}^3$ $\{ e_1,e_2,e_3\}$).The $2$-distribution $D_2:\mathbb{R}^3 \rightarrow T\mathbb{R}^3; (x,y,z) \mapsto ker(\omega(x,y,z))$ is integrable?

The kernel is the sets of the elements $(x,y,z)$ that satisfy $\omega(x,y,z)=0$, then:
$\begin{equation}
\begin{aligned}
\omega(x,y,z) &= x\cdot dx(x,y,z) + y\cdot dy(x,y,z) + z\cdot dz(x,y,z) \\
& = x\cdot dx(e_1+e_2+e_3)+y\cdot dy(e_1+e_2+e_3)+z\cdot dz(e_1+e_2+e_3) \\
& = x+y+z
\end{aligned}
\end{equation}$
The equality $\omega(x,y,z)=0$ give a set of vectors generated by $\{(1,0,-1),(1,-1,0)\}$, then $ker(\omega(x,y,z))=span\{(1,0,-1),(1,-1,0)\}$. 
So, the vector fields associated are $X=dx-dz, Y=dx-dy$.
I need to see if the Lie bracket is involutive, so using the Frobenius theorem i can show that $D_2$ is integrable
$\begin{equation}
\begin{aligned}
\space[ X,Y ] &= X(1)dx-X(1)dy-Y(1)dx+Y(1)dz \\
& = 0 
\end{aligned}
\end{equation}$
Since the Lie bracket is $0$ then $D_2$ is involutive, using the Frobenius theorem, then $D_2$ is integrable.
It's right my solution?
 A: Note that $\omega = d\big(\frac12(x^2+y^2+z^2)\big)$, so the integral manifolds of $\omega = 0$ are the spheres $x^2+y^2+z^2=c$.
A: Strictly speaking, $\omega$ does not define a $2$-plane distribution: Since $\omega_{(0, 0, 0)} = 0$, the subspace $(D_2)_{(0, 0, 0)}$ is all of $T_{(0, 0, 0)} \Bbb R^3$, which has dimension $3$. We ignore this issue by instead treating $D_2$ as the two plane distribution $\ker \omega \vert_{\Bbb R^3 \setminus \{(0, 0, 0)\}}$ on $\Bbb R^3 \setminus \{(0, 0, 0)\}$.
Hint The distribution $D_2$ is given as the kernel $\ker \omega$ of a $1$-form, which suggests using the form characterization of involutive distributions, rather than the dual vector field formulation---which requires computing local frames for $\ker \omega$ on each chart in some suitable cover of the manifold.
For a hypersurface distribution (i.e., the kernel $\ker \omega$ of a single nonvanishing $1$-form) this condition simplifies to the condition that $$d\omega = \alpha \wedge \omega$$ for some $1$-form $\alpha$ (at least locally), or equivalently that $$\omega \wedge d\omega = 0 .$$

 But computing gives that $$d\omega = d(x \,dx + \cdots) = dx \wedge dx + \cdots = 0 .$$ Thus the distribution is integrable.

A: No, your solution is not correct. For a vector $v = v_x\partial_x + v_y\partial_y+v_z\partial_z$, $v \in \ker \omega \iff xv_x+yv_y+zv_z=0$. This gives a two-dimensional distribution $D_2 = \ker \omega$ everywhere except at the origin (where the distribution becomes three-dimensional). To solve the problem via your method, you would need to find a basis of vector fields that span $D_2$ at each point. You will require different bases in different regions. For example, for $x\ne 0$, a suitable basis is $\lbrace y\partial_x - x\partial_y, z\partial_x - x\partial_y\rbrace$ - note these are both non-zero for $x\ne 0$. However the first vanishes along the $z$-axis, while the second vanishes along the $y$-axis, which is why you need to consider different regions/cases.
Alternatively, you could use Ted's answer to jump straight to the integral manifolds, proving integrability more directly. It's probably worthwhile for you to think about how these two approaches relate to each other. 
A: Here's an alternative solution (equivalent to the ones presented in the other answers), in terms of vector calculus: a nowhere-vanishing vector field $N$ along some open subset of $\mathbb{R}^3$ defines there the orthogonal distribution $N^\perp$. The distribution $N^\perp$ is integrable if and only if $N$ and ${\rm curl}\,N$ are orthogonal at every point. In your case, we recognize $$N = x\partial_x+y\partial_y+z\partial_z$$by dualizing $\omega$. One readily computes ${\rm curl}\,N = 0$, so we are done. This corresponds to the fact that $\omega$ is closed (and since $\mathbb{R}^3$ is simply-connected, actually exact, as shown in Ted's answer).
