Suppose we have two lines given by $a_{1}x + b_{1}y + c_{1}z = 0$ and $a_{2}x + b_{2}y + c_{3}z = 0$, where $(x,y,z) \in \mathbf{C^3}$.
What I am confused about, is that clearly $(0,0,0)$ is a point on both lines.
However if these two lines are distinct, then I should think that this is the only point of intersection, since I keep hearing people say "two distinct lines intersect at most at a point".
However, these two lines must intersect in the complex projective space $\mathbf{P^2}$; either they interect when $z = 1$ in $\mathbf{C^2}$, or at the line at infinity when $z = 0$. And if they intersect at a point $(a,b,c)$ then they also intersect at all nonzero complex scalar multiples of this point.
How do I reconcile the fact that I've been hearing for years that two distinct lines intersect at most at one point, with the fact that we know there are (a whole scalar multiples) of points with coordinates not all zero, that lie on both lines.
The projective plane it defined as $$ \mathbb{P}^2 = \frac{\mathbb{C}^3 \backslash \{(0,0,0)\} }{\sim} $$ where $(a,b,c) \sim (d,e,f)$ if there exists $\lambda \in \mathbb{C}^\ast$ such that $(a,b,c) = (\lambda d,\lambda e,\lambda f)$. Hence a point $(a:b:c)$ is an equivalence class that encodes the vector $(a,b,c)\in \mathbb{C}^3$ and all it's nonzero multiples i.e. a line through the origin (except for the origin itself).
As you know, a line in $\mathbb{P}^2$ has a homogeneous equation and the intersection of two (distinct) such lines can be described by the solutions of a linear system $$ \left\{ \begin{matrix} a_1x+b_1y+c_1z =0 \\ a_2x+b_2y+c_2z =0 \end{matrix}\right., $$ which in turn gives you a line $L\subset \mathbb{C}^3$ through the origin, since the system above is homogeneous of rank $2$. Therefore the quotient $(L\backslash \{(0,0,0)\})/\sim$ gives you a point in $\mathbb{P}^2$.
Let me know if it's clear...