I solved this problem on my own, months ago, but the solution seems to me completely forgotten, a little help on it would be appreciated:
Suppose $\alpha= \alpha(t)$ on an interval $I$ is a smooth (of class C$^1$) parametric representation of the curve $C$, and for any $t \in I$ we have $\space\space\frac{d}{dt}\alpha(t).v=0$, where $v$ is a constant vector, furthermore $\alpha(0)$ is perpendicular to $v$. Then $\forall t\in I: \space \alpha(t).v=0$.
On
Let $ \; t_0 \in \mathrm{dom(\alpha)} $ the FTC for vector valued functions implies $$\require{cancel}\alpha(t_0 ) -\alpha(0) = \int_{0}^{t_0} \dot\alpha(v)dv\qquad(1)$$ Now apply dot product upon (1) and $v$ will result (using $<,>$ as the dot operator) $$<v,\alpha(t_0)> -\cancelto{0}{<v,\alpha(0)>}=<v,\int_{0}^{t_0}\dot \alpha(u)du>$$ But $$<v,\int_{0}^{t_0}\dot \alpha(u)du>=\int_{0}^{t_0}<v,\dot \alpha(u)>du$$ Using our hypothesis and the fact that $t_0$ was taken arbitrarily it follows $$\forall t\in\mathrm{dom}(\alpha) \quad<\alpha(t),v>=0$$ Q.E.D.
Assume the ambient space is $\Bbb R^n$. Then $\alpha'(t)$ always lives in $\text{ker} \,v$, a subspace orthogonal to $v$. This suggests we use an orthogonal transformation $\rho$ to change coordinates, so that $\rho w$ is in the same direction as $e_n$ and $\rho\alpha'(t)$ lives in the span of $e_1,\cdots,e_{n-1}$, denoted $V$. We now only have to show $\rho\alpha(t)$ lives in $V+h$ for some $h\in\Bbb R^n$. Denote $\beta:=\rho\alpha$ then $\beta'=\rho\alpha'$. We only look at the $n$-th componenent of $\beta$ now. I assume by $I$ you mean $[0,1]$. So let $h:=\beta(0)$, we claim $\beta(I)\subset h+V$: otherwise suppose $\exists t_0$ s.t. $\beta^n(t_0)-\beta^n(0)\ne 0$. However, by the fundamental theorem of calculus applied to $\beta^n$ we know that $$\beta^n(t_0)-\beta^n(0)=\int_{[0,t_0]}(\beta^n)'(t)\mathrm dt=\int 0 = 0.$$