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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

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    Is $x$ a given non-nul vector, or this relation is true for every $x$ ? Note that for instance in dim 4, if (a,b,c,x) is an orthonormal base then the relation is true, and (a,b,c) are not in the same plane. Can you repcise your question ? – 2017-02-07
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    @zwim Sorry, I thought I had written it but forgot to. this is true for all $x$ lying on some given plane. – 2017-02-07
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    Even though, the same issue resides, in dim 5 consider the plan orthogonal to the dim 3 orthonormal base. $a=(1,0,0,0,0)$ $b=(0,1,0,0,0)$ $c=(0,0,1,0,0)$ and $x=(0,0,0,\alpha,\beta)$. – 2017-02-07
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    @dxiv Thanks, can you post this as an answer so that the question does not remain unanswered? – 2017-02-07

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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.

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    This is only true in dim 3. – 2017-02-07
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    @zwim Right, that's what the post [linked](http://physics.stackexchange.com/q/81204) into the OP implies. Could have been spelled out better in the question here, I guess. – 2017-02-07
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    Yep, I'm sorry that I didn't point that out at first. – 2017-02-07
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    P.S. It also holds true in higher dimensional spaces if one reads "plane" as "hyperplane". – 2017-02-07