smooth curve that is tangent to a 1-form kernel in every point

Let $$α = dz - ydx \in Ω^1 (\mathbb{R}^3)$$.

Prove that $$\forall p,q \in \mathbb{R}^3,\ \exists \gamma: [0,1] \rightarrow \mathbb{R}^3$$ smooth, such that $$γ(0)=p, γ(1) =q$$ and $$\gamma$$ is tangent to $$ker\ \alpha$$ for all $$t\in [0,1]$$.

If we take $$\gamma(t) = p(1-t) + qt$$ we math the conditions $$γ(0)=p, γ(1) =q$$, but how to match the last condition : $$\alpha (\gamma'(t)) = 0$$?

Thank you for any insights.

The third condition says that for each $$t$$, we must have $$\alpha(\gamma(t)) [\gamma'(t)] = 0. \tag{*}$$ If we write $$\gamma(t) = (x(t), y(t), z(t))$$ then $$\gamma'(t) = (x'(t), y'(t), z'(t))$$ and equation (*) becomes $$z'(t) - y(t) x'(t) = 0,$$ which you can, perhaps, solve (or can at least prove has a solution).