Let $p$ be prime. We say that a geometry is an affine plane if it satisfies three properties:
(i) Any two distinct points determine a unique line.
(ii) There exists a set of four points so that no three of them are collinear.
(iii) The parallel postulate.
We say that it is projective if instead of (iii), it satisfies
(iii)' Any two lines have nonempty intersection.
Now consider $\mathbb Z_p^2$ and $\mathbb Z_p^3$ as geometries. Say lines are of the form $\{\vec x+t \vec y: t\in \mathbb Z_p \}$ where $\vec x, \vec y$ are in the relevant geometry with $\vec y$ nonzero.
I am having trouble seeing why $\mathbb Z_p^2$ satisfies properties (ii) and (iii). I am also not sure why $\mathbb Z_p^3$ fails to satisfy properties (iii) and (iii)'.
I was able to see by inspection why $\mathbb Z_3^2$ is an affine plane, but I'd really appreciate some help in these more general cases.
For this question, it's no different than why the parallel postulate holds in $\mathbb{R}^2$, but not in $\mathbb{R}^3$, or for that matter, in $K^2$, but not in $K^3$, where $K$ is an arbitrary field.
In $K^3$, let \begin{align*} &{\small{\bullet}}\;\,L\;\text{be the line}\;\{(t,0,0) \mid t \in K\}\\[4pt] &{\small{\bullet}}\;\,P\;\text{be the point}\;(0,0,1)\\[4pt] &{\small{\bullet}}\;\,A\;\text{be the line}\;\{(a,0,1) \mid a \in K\}\\[4pt] &{\small{\bullet}}\;\,B\;\text{be the line}\;\{(0,b,1) \mid b \in K\}\\[4pt] \end{align*} Then $A,B$ are two distinct lines through $P$, neither of which meets $L$.
Next, consider $K^2$.
It's easy to verify that the $4$ points $$(0,0),\;\;(1,0),\;\;(1,1),\;\;(0,1)$$ are such that no three are collinear. One way to check this is to find, for each pair of the $4$ points, the vector equation of the line they determine, and then verify that the other two points fail to satisfy the equation. I'll leave that for you check.
To show the parallel postulate holds in $K^2$, let $L\;\text{be the line}\;\{\vec x+t\vec y \mid t \in K\}$, for some $\vec y \in K^2,\;\vec y \ne \vec 0$, and let $\vec q$ be a point of $K^2$ not on $L$.
We need to show
Let $A$ be the line $\{\vec q + a\vec y \mid a \in K\}$.
If $A$ meets $L$, then \begin{align*} &\vec q + a\vec y = \vec x + t\vec y\;\text{for some}\;a,t \in K\\[4pt] \implies\;&\vec q = \vec x + (t-a)\vec y\\[4pt] \implies\;&\vec q \in L\\[4pt] \end{align*} contrary to the choice of $\vec q$.
Thus, $A$ is one line through $\vec q$ which doesn't meet $L$.
Next, suppose $B$ is any line through $q$ which doesn't meet $L$.
Since $B$ goes through $q$, we can write $B = \{\vec q + b\vec z \mid b \in K\}$, for some $\vec z \in K^2.\;\vec z \ne \vec 0$.
Since $B$ doesn't meet $L$, there do not exist $b,t \in K$ such that $$\vec q + b\vec z = x + t\vec y$$ Equivalently, in terms of coordinates, the system of two equations \begin{align*} q_1 + bz_1 &= x_1 + ty_1\\[4pt] q_2 + bz_2 &= x_2 + ty_2\\[4pt] \end{align*} in the two unknowns $b,t$ has no solution in $K$.
Solving via elementary algebra, the inconsistency of the system yields $$z_1y_2 = z_2y_1$$ hence, since the vectors $\vec z,\vec y$ are nonzero, each must be a nonzero multiple of the other.
It follows that the lines $A,B$ are identical (they have the same initial point, $\vec q$, and parallel, nonzero direction vectors).
Thus, $K^2$ satisfies the parallel postulate, as was to be shown.