You can define the projective plane $P^2$ as the space of lines in $\mathbb{R}^3$ through the origin. In particular, given a line in $\mathbb{R}^3$ which passes through $(0, 0, 0)$, there is a corresponding point in the projective plane $P^2$.
Suppose $\ell$ is a line in $\mathbb{R}^3$ through the origin. Any line is uniquely determined by two different points on the line. We know that $(0, 0, 0)$ is a point on the line, so suppose $(a, b, c) \neq (0, 0, 0)$ is a different point on $\ell$. The point of $P^2$ corresponding to the line $\ell$ in $\mathbb{R}^3$ is denoted by $[a, b, c]$. Note that we could have chosen a different point on the line $\ell$, but every other point is of the form $(ka, kb, kc)$ for some non-zero real number $k$, so we need $[a, b, c]$, the point in $P^2$ representing $\ell$, to be the same as $[ka, kb, kc]$ because it represents the same line. That is, we require $[a, b, c] = [ka, kb, kc]$ for all non-zero real numbers $k$. This is why we use homogeneous coordinates on $P^2$.
With this in mind, we can now understand the relationship between points in $\mathbb{R}^3$, other than the origin, and points in $P^2$. Given any point $(a, b, c) \neq (0, 0, 0)$ in $\mathbb{R}^3$, there is a corresponding point $[a, b, c]$ in $P^2$ which represents the line in $\mathbb{R}^3$ which passes through $(0, 0, 0)$ and $(a, b, c)$. The reason this correspondence fails for $(a, b, c) = (0, 0, 0)$ is that you don't have two different points in $\mathbb{R}^3$, so you don't get a line (and hence, you don't get a point in $P^2$).
Note that a point at infinity of $P^2$ is a point of the form $[a, b, 0]$, and corresponds to a line in $\mathbb{R}^3$ through $(0, 0, 0)$ and $(a, b, 0)$; in particular, it lies in the $xy$-plane. It follows that the line at infinity of $P^2$ corresponds to the lines in $\mathbb{R}^3$ through the origin which lie in the $xy$-plane.
