Suppose I have a hyperplane in $n\geq 3$ dimensional space, that does not pass through the origin. Then, in the case of a plane in three-space, the set of points on the plane which have a $0$ entry, e.g. $(3, 0, 1)$, $(0, 0, 9)$, all lie on a set of at most three lines which are contained in the plane.
In the general case for hyperplanes, does this same property hold? That is, given a hyperplane in $n>3$ dimensional space which does not pass through the origin, do the set of all points lying on that plane which have a $0$ entry lie on a finite set of lines contained in that hyperplane?
The set of points with a $0$ in the $n$th coordinate lie in a plane, $X_n$. So the set of points in your plane $A$ with a $0$ coordinate is $A \cap (\bigcup_n X_n) = \bigcup_n (A \cap X_n)$. the intersection of two $n-1$ dimensional planes is either empty, a $n-2$ dimensional plane, or a $n-1$ dimensional plane. The latter only occurs when the two planes are equal.
If $A$ doesn't go through the origin, then it can't be equal to any $X_n$, so the intersection is empty or a $n-2$ dimensional space contained in $A$ (the analogue of the lines in 3D). This means the set you're looking for is an intersection of $n$ or less $n-2$ dimensional spaces contained in $A$.