How to rotate a plane around a line that lies on it by alpha degrees? I am not much of a mathematician and I hope I could get some help with my problem.
I have two points in 3D and they form a line. Using 3D programming I am slicing the volume with a plane. All I need for creating a plane here is a single point and a normal vector of that plane.
The thing is there are infinite planes that could fit the line. All those planes actually circle the line. My idea is to determine the plane normal with an angle variable.
So what I'm asking is how can I determine a plane normal with a given angle variable?

EDIT: Just to clarify my sketch: black line is a line formed with two points I have and red vectors are plane normal vectors that I'd like to determine with an angle (phi in the picture).
Thank you for any given help, m.k.
 A: So, you have two points $A,B$ on the line, then you have the vector
$\vec {AB} $ and the unitary vector
$$
{\bf t} = {{\vec {AB} } \over {\left| {\vec {AB} } \right|}}
$$
Now, as far as I understood, you have a plane containing those two points
(and the line through them) that you consider as the reference plane, i.e.
the plane about which to measure the rotation angle.
I do not know how you are going to "fix" that plane. Seems that you know how to determine
the unit normal vector to it (let's call that ${\bf n}$).   
If not, let's suppose that you individuate such a reference plane by giving a third point ($C$)
contained in it.
Then take ${\vec {AC}}$ and the dot product $a = {\vec {AC}} \cdot {\bf t}$.
$a \, {\bf t}$ is the component of ${\vec {AC}}$  along ${\vec {AB}}$ , which when subtracted from ${\vec {AC}}$
$$
{\vec {AC'}}  = {\vec {AC}}  - a\;{\bf t}
$$
leaves the component orthogonal to ${\vec {AB}}$.
Normalize this to obtain 
$$
{\bf u} = {{\vec {AC'} } \over {\left| {\vec {AC'} } \right|}}
$$
which is normal to ${\bf t}$ and also in the reference plane.
Now 
$$
{\bf n} = \;{\bf t}\; \times \;{\bf u}
$$
is the unitary vector normal to the reference plane.
In case, change its sign, so to obtain the required direction for measuring the angle:
we are considering that the angle $\alpha$ starts from ${\bf u} \to \alpha =0$
and moves towards ${\bf n} \to \alpha =\pi /2$.
Therefore
$$
\left\{ \matrix{
  {\bf u}' = \cos \alpha \;{\bf u} + \sin \alpha \;{\bf n} \hfill \cr 
  {\bf n}' =  - \sin \alpha \;{\bf u} + \cos \alpha \;{\bf n} \hfill \cr}  \right.
$$
where  $ {\bf u}' $ lies on the rotated plane, $ {\bf n}' $ is normal to that, and both
are orthogonal to $ {\bf t}$.   
What you are asking is $ {\bf n}' $.
