= Solution
The differential-forms version of the <Frobenius theorem> says that a constant-rank <distribution (differential geometry)> $D=\bigcap_i\ker\alpha^i$ is <integrable distribution>[integrable] exactly when
$$
d\alpha^i=\sum_j\beta^i{}_j\wedge\alpha^j
$$
for suitable 1-forms $\beta^i{}_j$. For a <plane distribution> $D=\ker\alpha$ on $\mathbb R^3$, this reduces to the <integrability criterion for a plane distribution> $\alpha\wedge d\alpha=0$.
For example, $\ker dz$ is integrable: its integral surfaces are the horizontal planes $z=\text{constant}$. In contrast, for $\alpha=dz-x\,dy$,
$$
\alpha\wedge d\alpha=(dz-x\,dy)\wedge(-dx\wedge dy)=-dz\wedge dx\wedge dy\ne0.
$$
Thus $\ker(dz-x\,dy)$ is not integrable; it is the standard <contact structure> on $\mathbb R^3$.
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