Relative position of three planes

To study the relative position of three planes $π_{1} (A; \vec{u}, \vec{v}), π_{2} (A^{'}; \vec{u^{'}}, \vec{v^{'}})$ and $π_{3} (A^{″}; \vec{u^{″}}, \vec{v^{″}})$ expressed by their general equations:

$\begin{array}{rrcl} π_{1} : & A_{1} x + B_{1} y + C_{1} z + D_{1} & = & 0 \\ π_{2} : & A_{2} x + B_{2} y + C_{2} z + D_{2} & = & 0 \\ π_{3} : & A_{3} x + B_{3} y + C_{3} z + D_{3} & = & 0 \end{array}$

Let's consider the system formed by three equations. The matrix $M$ and $M^{'}$ associated with the system are: $M = (\begin{matrix} A_{1} & B_{1} & C_{1} \\ A_{2} & B_{2} & C_{2} \\ A_{3} & B_{3} & C_{3} \end{matrix})$ $M^{'} = (\begin{matrix} A_{1} & B_{1} & C_{1} & - D_{1} \\ A_{2} & B_{2} & C_{2} & - D_{2} \\ A_{3} & B_{3} & C_{3} & - D_{3} \end{matrix})$

We can classify the relative position of the planes by the compatibility of the systems:

Compatible system
- Indeterminate Compatible system
  - $r a n k (M) = r a n k (M^{'}) = 1$
    
    The solutions depend on two parameters. Three planes are equal.
  - $r a n k (M) = r a n k (M^{'}) = 2$
    
    The solutions depend on a parameter, therefore they have a straight line in common.
    
    Now we must determine the position of the planes two by two. We have 2 options:
    - Three planes are cutting in a straight line.
    - Two planes are equal and cut the other plane.
- Determinate compatible system
  - $r a n k (M) = r a n k (M^{'}) = 3$
    
    The planes are cutting at a point.
- Incompatible system
  - $r a n k (M) = 1$ ; $r a n k (M^{'}) = 2$
    
    Incompatible system: parallel planes.
    
    Next we must determine if they coincide or not.
  - $r a n k (M) = 2$ ; $r a n k (M^{'}) = 3$
    
    Incompatible system: they are cutting planes.
    
    Next, it is necessary to determine if there are also parallel planes.