The normal of a surface that passes through the origin this is my first question here. 
Suppose I have a surface as follow:
$$x^2+y^2+z^2=9$$
The gradient of the surface at a particular point $P = (x_0,y_0,z_0)$ is just $(2x_0,2y_0,2z_0)$ and the parametric equation of the line is $x(t) = x_0 + t(2x_0)$ and so on. Now how can we find the points that have their normal passing through the origin?
Normally I would set $P = (0,0,0)$ and the direction vector would remain. But here the direction vector is directly dependent on the point chosen, so if I were to set $P = (0,0,0)$, then there would be no line as well. 
Could you guys help me with this? 
Thanks 


how can we find the points that have their normal passing through the origin?

Let $M(t)= (x_0 +2x_0 t,y_0 +2y_0 t,z0 +2z_0 t )$.
Check if the equation $M(t) = (0,0,0)$ has a solution.
In you initial problem every point has their normal passing through the origin, because for every point the equation has a solution ($t=-1/2$).
