Normal planes to a curve passing through a line Show that the normal planes to the curve 
$$x=a \cos t$$
$$y=a \sin\alpha \sin t$$
$$z=a \cos\alpha \sin t$$
are passing through the straight line $x=0$; $z + y \tan\alpha = 0$.

As I see now that you have worked on the subject, here is a solution:
The normal plane at epoch $t_0$ is the plane having as normal vector a tangent vector to the curve, 


*

*i.e., taking the derivatives, with equation 


$$\tag{1}x(-a \sin t_0)+y(a \sin \alpha \cos t_0)+z(a \cos \alpha \cos t_0)=h.$$


*

*with $h$ determined by the fact that this plane passes through point 


$$\tag{2}(x_0=a\cos t_0 , y_0=a \sin \alpha \sin t_0 , z_0=a \cos \alpha \sin t_0)$$
which means that, replacing in (1) $(x,y,z)$ with $(x_0,y_0,z_0)$ given by (2), we get:
$$\tag{3}h= \left[-1+(\sin \alpha)^2 +(\cos \alpha)^2\right)]a^2\cos t_0\sin t_0=0.$$
Thus this is a plane passing through the origin.
It remains to show that $(x,y,z) \in $ (L) (the line) $\implies (x,y,z) \in $ (P) the plane (recall that implication between characteristic properties corresponds to set inclusion), i.e.,
$$\begin{cases}x=0\\ z = - y \tan \alpha \end{cases} \ \ \ \implies x(-a \sin t_0)+y(a \sin \alpha \cos t_0)+z(a \cos \alpha \cos t_0)=0.$$
which is almost immediately established.
