Let $\alpha\colon I\rightarrow \mathbb{R^3}$ be a regular curve parametrized by arc length, such as $\kappa (s)>0,\,\forall s\in I$. Show that: if all the osculating planes of $\alpha$ pass through a fixed point, then $\alpha$ is a plane curve.
My attempt:
We know the osculating plane is normal to the binormal vector B(s).
Let $R(x_0,y_0,z_0)$ be the common intersection of the osculating planes. Then the equation of the osculating plane at any point $P(a,b,c)$ of the curve is given by $$\langle(R-P),B(s)\rangle=0, \,\forall s\in I$$ Differentiating, we get \begin{align} \langle(R-P)',B(s)\rangle+\langle(R-P),B'(s)\rangle &=0\\ \langle(R-P),B'(s)\rangle &=0\\ \langle(R-P),\tau(s)N(s)\rangle &=0. \end{align}
Now, suppose $\alpha$ is not a plane curve, that is, $\tau \neq 0$. Thus, we have $\langle(R-P),N(s)\rangle =0$, that is, $(R-P)$ is orthogonal to $N(s)$. But also, $\langle(R-P),B(s)\rangle=0$, hence $(R-P)$ is paralell to $T(s)$.
However, I don't see how I can conclude the proof from there. What should I do next? Also, is my writing correct?
I appreciate any help!