Suppose $\sigma:[a,b] \to \mathbb{R}^3$ and is of class at least $C^2$. Suppose for all $t \in [a,b]$ we have that $$\dim \operatorname{Span}_{\mathbb{R}}\{\sigma(t),\sigma'(t),\sigma''(t)\}=2$$ and that $\sigma(t)$ and $\sigma'(t)$ are linearly independent. How can I prove that there exists a plane $\Pi$ such that $\sigma([a,b]) \subset \Pi$?
My attempt:
I tried computing the binormal vector $b(t)=\frac{\sigma'(t) \wedge \sigma''(t)}{\|\sigma'(t) \wedge \sigma''(t)\|}=\frac{a(t)}{|a(t)|}\frac{\sigma'(t) \wedge \sigma(t)}{\|\sigma'(t) \wedge \sigma(t)\|}$ where $a(t)$ is such that $\sigma''(t)=a(t)\sigma(t)+c(t)\sigma'(t)$.
Hint:
Given what you've done so far, \begin{align} \frac{d}{dt}(\sigma(t)\wedge\sigma'(t))&=\sigma'(t)\wedge\sigma'(t)+\sigma(t)\wedge\sigma''(t)\\ &=c(t)(\sigma(t)\wedge\sigma'(t))\\ \implies \sigma(t)\wedge\sigma'(t)&=e^{\int_0^tc(s)ds}(\sigma(0)\wedge\sigma'(0))\\ \implies \frac{(\sigma(t)\wedge\sigma'(t))}{\| (\sigma(t)\wedge\sigma'(t))\|}&= \frac{(\sigma(0)\wedge\sigma'(0))}{\| (\sigma(0)\wedge\sigma'(0))\|}\ \text{ for all }\ t\ge0\ . \end{align}