Curve in 3-space has tangent always intersecting a fixed line, then the curve is planar This is an old exam question that I don't know how to get started.

Let $\alpha$ be a regular curve in $\mathbb{R}^2$. Prove that if any tangent line to $\alpha$ intersects a fixed line $l\in\mathbb{R}^3$, then $\alpha$ is planar.

My first problem is how to actually use the main assumption. I want to algebraically express the condition that these two lines intersect to derive that the torsion is $0$. I parameterize $l = a + b t$, but then I don't know to get an expression to differentiate (or manipulate) that characterizes the assumption that the tangent and $l$ intersect. Any hints?
 A: Let $\alpha\colon I \to \mathbb{R}^3$ be a unit speed parametrization of the regular curve. The tangent line at a point $\alpha(s)$ is $\lambda \mapsto \alpha(s)+\lambda T(s)$. Every tangent intersects the fixed line $l$, so for every $s\in I$ there is a $\lambda(s)$ such that
$$
  \beta(s) = \alpha(s) + \lambda(s) T(s) 
$$
lies on $l$.
At a point $s_0\in I$ three cases can occur:
Case 1: $\beta$ is regular in an open neighborhood around $s_0$. This means $\beta$ parametrizes (a piece of) the line $l$. Prove that $\alpha$ lies in a plane. This is the generic case. 
Case 2: $\beta$ is non-regular in an open neighborhood around $s_0$, i.e. $\beta(s)$ is a point. In this case you can prove that $\alpha$ is a particular plane curve.
Case 3: $\beta'(s_0)=0$ only at $s_0$. No need to prove anything in this case, but you should be aware these points can occur.
After considering these cases you have shown that the curve piecewise consists of planar curves. You should give an argument why the whole curve lies in the same plane.
