Suppose that your eye is at the origin, and the "canvas" on which you draw is in front of your eye, and is the $z = 1$ plane. Then the point $(2, 3, 5)$ in space will project, in your "drawing" or "seeing" of the world, to the point $(2/5, 3/5, 1)$ in the $z = 1$ plane. In general, any point $(x, y, z)$ will project to $(x/z, y/z, 1)$.
Now consider two parallel lines; the first one consisting of all points of the form $(1, t, 4)$ and the second consisting of points $(3, t, 2)$. The projections of these to the drawing plane will consist of points of the form $(1/4, t/4, 1)$ and $(3/2, t/2, 1)$, respectively, i.e., they'll still be parallel.
Now look at the lines $(-1, 0, t)$ and $(1, 0, t)$. These project to
$(-1/t, 0, 1)$ and $(1/t, 0, 1)$, which "meet" at the point $(0, 0, 1)$ when $t$ goes to $\infty$.
Does that help at all?
$\newcommand{\Reals}{\mathbf{R}}$Here's a "parametric" way to think of it: When you write "last coordinate $0$", presumably you're thinking of $\Reals^{3}$ embedded in $\Reals^{4}$ as $(x, y, z, 1)$. Take a non-zero direction $v = (a, b, c)$ in $\Reals^{3}$. A pair of parallel lines in $\Reals^{3}$ can be parametrized by
\begin{align*}
\ell_{1}: &\quad (x_{1} + at, y_{1} + bt, z_{1} + ct, 1)
\sim \tfrac{1}{t}(x_{1} + at, y_{1} + bt, z_{1} + ct, 1), \\
\ell_{2}: &\quad (x_{2} + at, y_{2} + bt, z_{2} + ct, 1)
\sim \tfrac{1}{t}(x_{2} + at, y_{2} + bt, z_{2} + ct, 1).
\end{align*}
Distribute the division by $t$, and let $|t| \to \infty$.
Best Answer
In standard coordinate geometry, a "vertical" line has equation of the form $x=a$. The homogeneous form of this is $X=aZ$. This line contains the point $(X:Y:Z)=(0:1:0)$ in the projective plane, no matter what $a$ is. So all "vertical" lines meet at $(0:1:0)$.