Suppose 3 planes intersect, 2 of the planes are parallel and the system is inconsistent Just like (b) in this picture what would the rank of the matrix be?
$A = \pmatrix{a_{11}&a_{12}&a_{13} \\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}}$
suppose the first and second equation for the plane are parrallel then
$A = \pmatrix{a_{11}&a_{12}&a_{13} \\ka_{11}&ka_{12}&ka_{13}\\a_{31}&a_{32}&a_{33}}$ where $k \in \Bbb{R}$ then we can apply elementary row operations to get
$A = \pmatrix{a_{11}&a_{12}&a_{13} \\0&0&0\\a_{31}&a_{32}&a_{33}}$ and since the whole system is inconsistent then $A = \pmatrix{a_{11}&a_{12}&a_{13} \\0&0&0\\0&0&0}$ and so the rank is 1. Is this a correct way of doing it?
Your logic up to here is fine. $A = \pmatrix{a_{11}&a_{12}&a_{13} \\0&0&0\\a_{31}&a_{32}&a_{33}}$
I am not sure on your reasoning to eliminate the 3rd row, though.
At this point you have two independent row vectors (because the two planes intersect) and one $\bf 0$ vector row.
That means you have a rank of $2$