If the osculating planes of a curve pass through a fixed point, the curve is a plane curve.
How to prove it?
[Math] the osculating planes of a curve pass through a fixed point $\rightarrow$ the curve is a plane curve.
differential-geometrygeometry
Related Solutions
Just apply Serret-Frenet.
WLOG the common point is the origin and the curve is parametrised by arclength $s$. The given condition says $\mathbf{r}\cdot(\mathbf{r}'\times\mathbf{r}'')=0$.
Since $\mathbf{r}'\times\mathbf{r}''=\mathbf{T}\times\mathbf{T}'=\kappa\mathbf{T}\times\mathbf{N}=\kappa\mathbf{B}$, either we have $\kappa=0$ on an interval (so we are done), or $\mathbf{r}\cdot\mathbf{B}=0$ on an interval.
Differentiating $\mathbf{r}\cdot\mathbf{B}=0$ gives $\mathbf{r}\cdot\tau\mathbf{N}=0$ so again either zero torsion on an interval (so we are done) or $\mathbf{r}\cdot\mathbf{N}=0$ on an interval. Since $\mathbf{r}\cdot\mathbf{N}=\mathbf{r}\cdot\mathbf{B}=0$ we have $\mathbf{r}=(\mathbf{r}\cdot\mathbf{T})\mathbf{T}$ so $\kappa=0$.
The osculating plane can be defined by the equation $\langle B,x\rangle=0$.
Let's assume the existence of some (constant) $p\in\mathbb R^3$ such that $\langle B,p\rangle\equiv0$. Since $p=0$ is a trivial solution for this, I'll also assume that $p\ne0$.
Taking the derivative we obtain $\langle B',p\rangle\equiv0$. Since $B'=\tau N$, we get $\tau\langle N,p\rangle\equiv0$.
Claim: $\tau\equiv0$. Suppose toward a contradiction that $\tau(s_0)\ne0$ for some $s_0$. Then $\langle N(s_0),p\rangle=0$. Taking derivatives, $$ \tau\langle N',p\rangle = \tau'\langle N,p\rangle + \tau\langle N',p\rangle = 0\quad(\textrm{at }s_0) $$ i.e., $$ \langle N',p\rangle = 0\quad(\textrm{at }s_0) $$ Substitute $N' = -kT-\tau B$ and get $$ k\langle T,p\rangle = k\langle T,p\rangle +\tau\langle B,p\rangle = 0\quad(\textrm{at }s_0) $$ Since $k\ne0$, it must be $\langle T(s_0),p\rangle$. Since $\langle N(s_0),p\rangle=0$ and $\langle B,p\rangle=0$ and $\{T, N, B\}$ is an orthonomal basis of $\mathbb R^3$, we get $p=0$; a contradiction. Then $\tau\equiv0$ as claimed.
Now we can use the result that says $\tau\equiv0\iff$im$(\alpha)$ is included in a plane. Since we are only interested here in the $\Rightarrow$ part, I'll recall how to prove it. Without loss of generality we will assume $\alpha(c)=0$ for some $c$ in the domain of $\alpha$.
First observe that $\tau\equiv0\implies B$ constant. Let $H$ be the osculating plane common to all points. As we saw above, $H$ is defined by the equation $\langle B,x\rangle=0$. Since $\langle B,T\rangle\equiv0$, the map $s\mapsto\langle B,\alpha\rangle$ is constant. And since it evaluates to $0$ at $c$, $$ \langle B,\alpha\rangle\equiv0, $$ i.e., $\alpha(s)\in H$ for all $s$.
Best Answer
Parametrize the curve by arclength, as usual, by $\alpha(s)$. Say the fixed point is the origin. Then there are functions $a(s)$ and $b(s)$ so that $$\alpha(s) = a(s)\mathbf T(s) + b(s)\mathbf N(s).$$ Now differentiate and apply Frenet.