$| A_r \cup A_s \cup A_t | = | A_r | + | A_s | + | A_t | - | A_r \cap A_s | - | A_s \cap A_t | - | A_t \cap A_r | + | A_r \cap A_s \cap A_t | $
For sets in condition 2 this gives $|A| \ge 3k - 9 + 2 \ge k + |A| + 1 - 9 + 2, k \le 6.$
if $k=6$, $|A| < 2k = 12$, also $|A| \ge 3k - 9 + 2 = 11$, $|A| = 11$.
For any pair$\{ a, b \}(a \ne b)$, it is contained in at least 3 subsets. Every subset gives $\tbinom{k}{2}$ pairs, hence $n \cdot \tbinom{k}{2} \ge 3 \tbinom{|A|}{2}$ which gives $n \ge 11$.
Consider pairs that contain the $i$-th element, let $a_i$ be the number of subsets containing the $i$-th element, then
$ \sum a_i = nk$,
$\sum_i \tbinom{a_i}{2} = \sum_{i<j} | A_i \cap A_j | \le 3 \tbinom{n}{2}$
Apply Cauchy's Inequality we get $n \le 11$.
Example for $n = 11$:
$A_i = \{ i, i+ 1, i + 2, i + 6, i + 8, i + 9\} (mod 11)$
element of $A$ can be viewed as point in unit circle, consider distance of two points in $A_i$, each possible distance $1,2,3,4,5$ appears exactly three times.