I have that
$A_1x + B_1y + C_1z = D_1$
$A_2x + B_2y + C_2z = D_2$
How do I intepret the solutions to these equations geometrically? What do they mean?
In $\mathbb R^3$, the solution set of a single linear equation is a plane (not necessarily through the origin). The solution set of a system of equations is the intersection of the solution sets of the equations in the system.
So...the solution set of a system of two linear equations is the intersection of two planes. This is either a plane (when the two equations describe the same plane), a line (if they are not the same plane and not parallel), or empty (if they are parallel).
These two equations represent two planes on the 3d space. These two planes might be either parallel (where there is no solution) or they do intersect. Their intersection will be either a straight line or a plane which means the two planes are identical.
There's a couple different ways.
Each equation can be considered a two dimensional plane in three dimensional space. The solutions $(x,y,z)$ to the equation are the points of the plane. Simultaneous solutions to the two equations will be where the planes intersect. Note this will generally happen in a line, but it could also not happen at all (if the planes are parallel and separated) or happen in a plane (if the planes coincide).
The $(A_1,A_2)$, $(B_1,B_2),$ $(C_1,C_2)$ and $(D_1,D_2)$ can be seen as two dimensional vectors and the two equations can be considered as $xA+yB+zC=D$ which says that $D$ is a linear combination of vectors $A,$ $B$ and $C.$ So the solutions are different ways to make the vector $D$ as a linear combination of $A$ $B$ and $C.$ Note that since the vectors are two-dimensional, as long as two of $A,B,C$ are linearly independent, you can always make $D$ as a linear combination(i.e. there is a solution). In this case, the third vector (if nonzero) will be redundant and this corresponds to the generic case above where the solution set forms a line (the one redundant vector is the one degree of freedom of the line). In the weird case where $A$ $B$ and $C$ are all parallel to one another and not linearly independent, then you can only make $D$ as a linear combination if it is parallel to them, and otherwise you can't (i.e. no solution). The first case here corresponds to the case above where the solution set forms a plane and the second to when there are no solutions (when the solutions to the individual equations were parallel, separated planes).