Using the definition of RANK here: rank of matrix
It is the maximum number of linearly independent row vectors.
We have that $\langle a_{11}, a_{12}, a_{13}\rangle$ is a normal vector for $P_1$ and likewise with the other column vectors.
Since these arent parallel, then why isn't the rank $ = 3$, since there are 3 linearly independent vectors?
Best Answer
You need to edit your title for this question. The three planes are distinctly not parallel. Three parallel planes would be like three floors of a building, one above the other, not meeting.
A simple example would be the three planes $z = 0 \ ; z = 1 \ ; z = 2 .$ In this case $$A = \pmatrix{0&0&1 \\0&0&1\\0&0&1}$$ which has rank 1.
The planes shown each intersect the others, pairwise, but there is no unique point of intersection, and no shared line of intersection, not like the above.
If rank A = 3 with no free variables, there would be a single unique point of intersection, so rank A is not 3. The simplest example to visualize this case it the three coordinate planes $z = 0 \ ; y = 0 \ ; x = 0 \ $. Here A is the identity matrix $$A = \pmatrix{1&0&0 \\0&1&0\\0&0&1}$$ and rank A = 3
If rank A = 2 there is one free variable so there would be a single one-dimensional intersection, a line. If the b's are all zero or otherwise consistent the planes would meet like the pages of a book. An example would be $ z = 0 \ ; y = 0 \ ; \ y + z = 0$
This gives the matrix $$A = \pmatrix{0&1&1 \\0&1&0\\0&0&1}$$
We can reduce this easily and we see rank A = 2
A change in the b's will make an inconsistent system so that the planes intersect pairwise as in the given diagram, but there is no point that is a solution to all three equations simultaneously.
The system $ z = 0 \ ; y = 0 \ y + z = 1$ will give a graph of the form given.
Therefore rank A = 2
Just to be complete: If rank A = 1, there are two free variables, so there is a two-dimensional solution, and in fact there is only one plane with the others coincident if the b's are zero or consistent, parallel planes otherwise