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I am trying to understand the question (this is Van Lint's Intro to Combinatorics) by first looking at when $r=2$. The points are ordered pairs with each entry in $\mathbb{F}_2$ and all possible linear combinations $x+ty$ where $x,y \in \mathbb{F}_2^2$ and $t\in\mathbb{F}_2$ make a tetrahedron.

My first question is what is a planes of $AG_2(2)$? It would also help to see what they are for $AG_3(2)$.

My confusion is, if $r$ is the number of elements in our field and $2$, the dimension is fixed, wouldn't there only be one plane regardless of the what $r$ we take? but the question mentions planes.

My second question is about what needs to be shown. From my understanding, what we want to do is take any $3$ points from the tetrahedron, and show that it belongs in exactly one block of size $4$ in the Steiner system given.

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  • $\begingroup$ I suppose $AG_r(q) $ stands for the affine geometric incidence structure over the $q$ element field, and plane means $2$ dimensional affine subspace. Then, $AG_2(q) $ has only one plane: the whole space. $\endgroup$ – Berci Mar 30 '18 at 22:24
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You have your $r$ and $q$ mixed up; $\mathrm{AG}_{r}(q)$ is affine geometry of dimension $r$ over $\mathbb{F}_{q}$ (commonly denoted $\mathrm{AG}(r,2)$). You only have one plane of $\mathrm{AG}_{2}(2)$, so it's not a very interesting case. When $r \geq 3$ you will have more (14 when $r=3$).

"what we want to do is take any 3 points from the tetrahedron": Not sure what tetrahedron you mean. You want to show if you take any 3 points they are contained together in precisely one block. This will follow from geometric arguments, I'm not sure what definitions and axioms are given by Van Lint but you will use the fact that 3 points cannot all lie on the same line.

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