Proof a (ALP) curve for which all osculating planes go through the origin is planar.
My approach: We write $$0 = \alpha(s)+ a(s)\mathbf T(s) + b(s)\mathbf N(s)$$ Then $$0=\mathbf{T}+a'\mathbf{T}+\kappa a\mathbf{N}+b'\mathbf{N}+b\mathbf{N}'$$ Using Frenet we conclude $$0=(1+a'-\kappa b)\mathbf{T}+(\kappa a+b')\mathbf{N}+\tau b\mathbf{B}$$ So then we conclude that $\tau b=0$, however I don't see how we can conclude that $\tau=0$ from here. For let's say $b(s_0)=0$, then we would have that $$0 =\alpha(s_0)+ a(s_0)\mathbf T(s_0)$$ $$0=(1+a')\mathbf{T}+(\kappa a+b')\mathbf{N}$$ i.e. $a'(s_0)=-1$ and $\kappa a+b'=0$, neither of which seem contradictory to anything...
You almost have the answer.
Your curve is differentiable, hence continuous. If $\tau(s_0)\neq 0$, then $\tau(s)\neq 0$, for $s$ in a neighborhood $U$ of $s_0$. Therefore, for $s\in U$, we have $b(s)=b'(s)=\kappa(s)a(s)=1+a'(s)=0$. We can then conclude that $\kappa(s)=0$, which means that the curve is on a straight line.