Spot it! and geometry over finite fields. Problem:

The party game “Spot It!” features 55 cards, each of which has eight symbols printed on it, in such a way that any two cards have exactly one symbol in common. (In the game, each player looks at a pair of cards and tries to find their common symbol.) The “Junior” version of the game has 30 cards with six symbols each. How do you use geometry over finite fields, as in the parts above, to build decks of cards with the required property? (Note: Spot It decks don’t quite have the optimal number of cards.)

My work:
This question is the third part of a problem. I already solved the two parts before where I prove the following formulas:
(I) The number of $m$-dimensional subspaces of an $n$-dimensional vector space over a field with order $q$ is $$\frac{(q^n - 1)(q^n - q)\ldots(q^n - q^{m-1})}{(q^m - 1)(q^m - q)\ldots(q^m - q^{m-1})}$$
(II) The number of two-dimensional spaces that contain a given 1-dimensional subspace is $$\frac{q^n - q}{q^2 - q} = \frac{q^{n-1} - 1}{q - 1}$$
So I am supposed to use those formulas or a similar method to solve the problem.
I already read this but none of the answers talks about the approach I am looking for, the most similar thing I was able to find is Example 4 of this answer but still doesn't solve my doubts.
I tried thinking of a field $F$ whose order $q$ is the number of symbols, then the set of all cards are all elements of $F^8$ where any two vectors only share a single coordinate or equivalently, a symbol. Since any two cards must share a single symbol or equivalently, a one-dimensional subspace of $F^8$ $$\frac{q^{7} - 1}{q - 1} = 2 \implies q^7 - 2q + 1= 0$$
but then I realised that the argument was wrong and didn't make much sense and that is why the equation yields wrong answers for the number of total symbols.
This section of the Wikipedia article about projective planes seems useful but I don't know how to apply it to the problem.
I would appreciate any hints on how to get started.
Thanks in advance.
 A: The intended solution here, using projective planes over finite fields, is to say that a card is a $1$-dimensional subspace of $\mathbb{F}_q^3$ for some $q$ and that a symbol on a card is a $2$-dimensional subspace containing a given $1$-dimensional subspace. Any two distinct $1$-dimensional subspaces span a unique $2$-dimensional subspace; this is what guarantees that any two cards have exactly one symbol in common. So the number of symbols on a card is the number of $2$-dimensional subspaces containing a given $1$-dimensional subspace, or
$$\frac{q^3 - q}{q^2 - q} = q + 1$$
as computed in part II. So for the original game we should take $q + 1 = 8$ and hence $q = 7$, and for the junior game we should take $q + 1 = 6$ and hence $q = 5$. The maximum number of cards allowed by this construction is the number of $1$-dimensional subspaces of $\mathbb{F}_q^3$, which is
$$\frac{q^3 - 1}{q - 1} = q^2 + q + 1$$
as computed in part I. So for the original game we can make up to $7^2 + 7 + 1 = 57$ cards and for the junior game we can make up to $5^2 + 5 + 1 = 31$ cards, hence the comments about the sizes of the decks not being optimal.
Previously this answer switched the $1$- and $2$-dimensional subspaces; this doesn't change anything because of projective duality.
If instead of $\mathbb{F}_q^3$ we used $\mathbb{F}_q^n$ for $n \ge 4$ then the number of symbols on a card would be $\frac{q^{n-1} - 1}{q - 1} = 1 + q + \dots + q^{n-2}$ which is a little too big to make equal to either $6$ or $8$.
