Geometric interpretation of ranks of matrices gathering coefficients of 3 affine and associated vectorial planes Given three planes $\pi_{1}, \pi_{2},\pi_{3}$ $\subseteq \mathbb{R}^{3}$ with their respect cartesian equation of the form $A_{i}x + B_{i}y + C_{i}z + D_{i} = 0$, we can determine their relative position in space by comparing the rank of the coefficient matrix $M$, with the rank of the augmented matrix $M^{*}$, where $$M = \begin{pmatrix}
A_{1}& B_{1} & C_{1} \\ 
 A_{2} & B_{2} & C_{2} \\ 
 A_{3}&  B_{3}& C_{3}
\end{pmatrix}$$ and $$ M^{*} = \begin{pmatrix}
A_{1}& B_{1} & C_{1} & D_{1}\\ 
 A_{2} & B_{2} & C_{2} & D_{2} \\ 
 A_{3}&  B_{3}& C_{3} & D_{3}
\end{pmatrix}$$ What is the geometric interpretation of these two matrices, how do they affect the planes relatively to each other visually? 
My guess has been that $rank(M)$ expresses if some of the planes have the same direction, while $rank(M^{*})$ also checks if they are separated by a distance or not. However, I have been struggling to understand why 3 separated parallel planes have $rank(M) = 1$ but $rank(M^{*}) = 2$, according to my textbook.

This array displays the different possible cases :

Let us first recall that the rank of a matrix $M$ can be considered at least from 2 angles of attack: (a) rank of its columns' space or (b) max. size of a non zero determinant of a submatrix.
Remark; there is a third characterization: (c) by using rank-nullity theorem: $rank=4-dim(ker(M))$.
Now, let us explain some aspects of the array above.
4 squares are hatched due to the fact that
$$rank(M^*) = \begin{cases} rank(M)\\ or \ \ rank(M)+1\end{cases}$$
Indeed,the rank of $M$ is (characterization (a) above) the dimension of $E:=vect (columns(M))$; extending $M$ to $M^*$ amounts to add a column belonging already to $E$ (rank preserved) or not (rank increased by one unit).
I do not give a detail for each case. Let me only give a treatment of the case you have been struggling with, the case of 3 parallel planes.
Their equations
$$ax+by+cz=d_k \ \ k=1,2,3$$
involve matrix :
$$\begin{pmatrix}
a&b&c&d_1\\ 
a&b&c&d_2\\  
a&b&c&d_3\\ 
\end{pmatrix}$$
on which we could discuss its rank (rank = 1 or rank = 2) using characterization (b). But it is simpler to consider it from the point of view of linear combinations:
$$\exists x,y,z \ s.t. \ x \begin{pmatrix}
a\\ 
a \\ 
a
\end{pmatrix}+y\begin{pmatrix}
b\\ 
b \\ 
b
\end{pmatrix}+z\begin{pmatrix}
c\\ 
c \\ 
c
\end{pmatrix}=\begin{pmatrix}
d_{1}\\ 
d_{2} \\ 
d_{3}
\end{pmatrix}\tag{2}$$
which will have a solution (in fact an infinite number of solutions) if and only if the last vector is in the range space of the first three ones, i.e., if and only if all its entries are identical. If one of the entries is different from the others, the rank of $M^*$ becomes $2$.
Let us come back a short while on the case where $d_1=d_2=d_3$. In this case, (2) is equivalent to the fact that $x+y+z-d_1=0$ has a 2 dimensional space of solutions.
