This is an illustration of the scenario I have:
I have the plane equation of a plane P $(A_px + B_py + C_pz + D_p = 0)$ and the coordinates of points A $(X_a, Y_a, Z_a)$ and B $(X_b, Y_b, Z_b)$. What I need is to find the equation of the plane Q, which is normal of the line segment AB. Is this possible, with the values I have?
$A_px + B_py + C_pz + D_p = 0$
has normal vector $\vec n= (A_p, B_p, C_p)$
$A (X_a, Y_a, Z_a)$ and $B (X_b, Y_b, Z_b)$
write vector $\vec {AB}=(X_b-X_a,Y_b-Y_a,Z_b-Z_a)$
the normal vector $\vec u$ of plane $Q$ is the cross product of the two
$\vec u=\vec n \times \vec {AB}=\\=(A_p (Z_b-Z_a)- C_p(Y_b-Y_a),C_p( X_b-X_a)-A_p (Z_b-Z_a),A_p(Y_b-Y_a)-B_p (X_b-X_a))$
plane $Q$ has equation
$\vec u_x\, x+\vec u_y \,y+\vec u_z\,z+d=0$
costant $d$ can be found substituting the coordinates of $A$ or $B$ in the previous equation
hope this helps