Is it possible to find a spatial curve $\alpha(s) = (x(s), y(s), z(s))$ such that the lines containing the normal vector $\hat{n}(s)$ at each point on the curve $\alpha(s)$ intersect the $z$-axis at a constant angle $\theta$?
My attempts: For any point $(r(s) \cos \phi(s), r(s) \sin \phi(s), h(s))$, the unit normal vector is given by $$ \hat{n}(s) = (\sin \theta \cos \phi(s), \sin \theta \sin \phi(s), \cos \theta). $$ The normal curvature vector is defined as $$ n(s) = \kappa(s) (\sin \theta \cos \phi(s), \sin \theta \sin \phi(s), \cos \theta). $$ with $\kappa(s)$ the curvature. Since the derivative of the unit tangent vector $\hat{T}(s)$ with respect to arc length $s$ is equal to the normal curvature vector, we have $$ \frac{d \hat{T}(s)}{ds} = \kappa(s) {\hat n}(s). $$ Therefore, the unit tangent vector $\hat{T}(s)$ can be expressed as $$ \hat{T}(s) = \left(\sin \theta \int_0^s \kappa(s')\cos \phi(s') \, ds', \, \sin \theta \int_0^s \kappa(s')\sin \phi(s') \, ds', \, \cos \theta \int_0^s \kappa(s') \, ds'\right). $$ Given that $\hat{T}(s)$ is a unit vector, it must satisfy the condition $$ \sin^2 \theta \left( \left( \int_0^s \kappa(s')\cos \phi(s') \, ds' \right)^2 + \left( \int_0^s \kappa(s')\sin \phi(s') \, ds' \right)^2 \right) + \cos^2 \theta \left( \int_0^s \kappa(s') \, ds' \right)^2 = 1. $$
Question: How can I find a solution to the above equations?