$\vec{a}\cdot\vec{x} = \vec{b}\cdot\vec{x} = \vec{c}\cdot\vec{x} = ct.$ then $\vec{a},\vec{b},\vec{c}$ lie on the same plane.

Question

I want to show that the vectors $\vec{a},\vec{b},\vec{c}$ lie on the same plane. The only property that I know they satisfy is

$$\vec{a}\cdot\vec{x} = \vec{b}\cdot\vec{x} = \vec{c}\cdot\vec{x} = ct,$$

where the $\vec{x}$ are the points of a plane. I know that $\vec{a}\cdot\vec{x}=ct.$ defines a plane, but can't get from here to where I want to go.

Thanks for any hint. To get more context on the question, please see the first answer in, where my abc vectors refer to the k vectors in that question:

Mattia (https://physics.stackexchange.com/users/31183/mattia), Show that the plane of incidence is perpendicular to the surface of reflection, URL (version: 2013-10-19): https://physics.stackexchange.com/q/81204

Answer 1

Hint: if $(\vec a - \vec b) \cdot \vec x = 0$ for all $\vec x$ in a plane, then $\vec a - \vec b$ is normal to that plane.

Therefore $\vec a - \vec b = \lambda \vec u\,$, and by symmetry also $\vec b - \vec c = \mu \vec u$ where $\vec u$ is a normal to the given plane, then the linear dependency between $\vec a , \vec b, \vec c\,$ follows easily.