The example in $\mathbb{R}^3$ given by $$ \omega = x\,\mathrm{d}y\wedge\mathrm{d}z + y\,\mathrm{d}z\wedge\mathrm{d}x + z\,\mathrm{d}x\wedge\mathrm{d}y, $$ shows that the kernel of $\omega$ need not be a line bundle. Away from the origin, the kernel is spanned by the radial vector field $$ R = x\,\frac{\partial\ }{\partial x} + y\,\frac{\partial\ }{\partial y} + z\,\frac{\partial\ }{\partial z}, $$ and there cannot be a continuous nonvanishing vector field that is a multiple of $R$ everywhere except at zero.

The example in $\mathbb{R}^3$ given by $$ \omega = x\,\mathrm{d}y\wedge\mathrm{d}z + y\,\mathrm{d}z\wedge\mathrm{d}x + z\,\mathrm{d}x\wedge\mathrm{d}y, $$ shows that the kernel of $\omega$ need not be a line bundle. Away from the origin, the kernel is spanned by the radial vector field $$ R = x\,\frac{\partial\ }{\partial x} + y\,\frac{\partial\ }{\partial y} + z\,\frac{\partial\ }{\partial z}, $$ and there cannot be a continuous nonvanishing vector field that is a multiple of $R$ everywhere except at the origin.

A similar argument works in the case of the closed $2$-form $$ \omega = x\,\mathrm{d}y\wedge\mathrm{d}z + y\,\mathrm{d}z\wedge\mathrm{d}x - 2z\,\mathrm{d}x\wedge\mathrm{d}y. $$