# intersection of three planes different cases, algebraic and geometric explanations

http://www.vitutor.com/geometry/space/three_planes.html

Could anyone help me to understand the following cases

$1.$ CASE (2.1), why rank of co-ef matrix is $2$ and augmented matrix is $3$? I can explain from the fact that if rank of co-ef matrix is $1$ and rank of augmented matrix is $3$ then it is case that all the three planes are parallel. And if rank of both is $3$ then it is case $1$, so what is the geometric explanation of case $2.1$?

$2.$ case $4.1$, shouldn't it be rank of augmented matrix is $3$? because three planes have different distances from the origin?

$3.$ case $3.1$ I also want the geometric and algebraic explanation.

Some of the ideas:

Generically (which to a mathematician means something like "most of the time"), any two planes will intersect at a line. In fact two planes either intersect at a line or are parallel and we know that two rows have to be the same (in the non-augmented matrix) for the two planes described by those rows to be parallel.

The rank $r$ is the number of planes that are linearly independent (in a sense). You have to try imagine adding planes together to get another plane in a way very similar to how you add vectors.

Then only if $r=3$ will you have a unique solution. This is the case where all three planes are linearly independent (again, in a sense).

This also should help you understand that if $r=1$ then all $3$ planes will be parallel.

If $r'>r$ then there will be no solution. I can't think of a geometric explanation (at least one in terms of planes) for this, but algebraically its because the target vector $b$ is outside the image of the transformation $x\mapsto Ax$.

Also, algebraically, $r'$ can't be more than $1$ higher than $r$ because you're only adding one more column. That column might be linearly independent of the other columns ($r'=r+1$) or linearly dependent on them ($r'=r$) but those are the only choices.