$\alpha$ is a plane curve if and only if all its osculator planes intersect at one point Let $\alpha$ be a regular curve. Prove that $\alpha$ is plane if and only if all the osculator planes intersect at one point.
I know that $\alpha$ is plane iff the binormal vector is constant, or iff the osculator plane is the same at every point. However, I don't know how to prove that.
NOTE: Originally, the statement said that "a curve is plane if and only if all the tangent planes intersect at one point". I understant that the tangent plane is the osculator plane, right?

HINT: Say all the osculating planes pass through the origin. This means that $$\alpha(s)=\lambda(s)T(s)+\mu(s)N(s)$$
for some functions $\lambda$, $\mu$. Now differentiate and use Frenet.
