Suppose I have an point $(3, 2)$ which I've seen has homogeneous coordinates $(3,2,1)$ and $k(3,2,1)$ where $k \neq 0$, and $(3,2,1)$ and $(3k,2k,k)$ both represent the same Cartesian point $(3, 2)$.
My question is that could I write $(3,2)$ in homogeneous coordinates as $(3,2,\frac{1}{2})$, $(3,2,\frac{1}{3})$, $\dots$ or equivalently $(6,4,1)$, $(9,6,1)$, $\dots$?
If so, are the homogeneous points $(6,4,1)$, $(9,6,1)$, $\dots$ the same as homogeneous points $(3,2,1)$, $(6,4,2)$, $(9,6,3)$, $\dots$, and hence $(6,4,2)$, $(9,6,3)$, $\dots$ are homogeneous coordinates of the Cartesian point $(3,2)$?
My last question is I need to see the sequence of points how $(3,2,1)$ approach to $(3,2,0)$?
I will use square brackets to denote homogeneous coordinates (this is standard notation), so $[a, b, c] = [ka, kb, kc]$ for all $k \neq 0$.
As you said, given a point $(a, b)$ in the plane, it has homogeneous coordinates $[a, b, 1]$, and a point with homogeneous coordinates $[a, b, 1]$ corresponds to the point $(a, b)$ in the Cartesian plane. Given a point with homogeneous coordinates $[a, b, c]$, then there are two cases to consider:
For example, the point $(3, 2)$ has homogeneous coordinates $[3, 2, 1]$, but this is also equal to $[6, 4, 2]$, $[9, 6, 3]$, etc. On the other hand, $[3, 2, \frac{1}{2}]$ is equal to $[6, 4, 1]$ and hence corresponds to the point $(6, 4)$, not $(3, 2)$.
Finally, note that the line through the origin and a point $(a, b) \neq (0, 0)$ can be parameterised by $(at, bt)$ for $t \in \mathbb{R}$, which is given by $[at, bt, 1]$ in homogeneous coordinates. For $t \neq 0$, we have $[at, bt, 1] = [a, b, \frac{1}{t}]$. As $\lim\limits_{t\to\pm\infty}\frac{1}{t} = 0$, we see that the line through the origin and the point $(a, b)$ meets the line at infinity at the point $[a, b, 0]$.