7・63 Let the set $A=\{0,1,2, \cdots, 9\},\left\{B_{1}, B_{2}, \cdots, B_{k}\right\}$ be a family of non-empty subsets of $A$, such that when $i \neq j$, $B_{i} \cap B_{j}$ has at most two elements. Find the maximum value of $k$.


[Solution] First, it is easy to see that the family of all subsets of $A$ containing at most 3 elements satisfies the requirements of the problem, where the number of subsets is
$$
C_{10}^{1}+C_{10}^{2}+C_{10}^{3}=175 \text {. }
$$

From this, we know that the maximum value of $k$ is no less than 175.
On the other hand, suppose there is a family of subsets $C$ that meets the requirements of the problem and contains a subset $B \in C$ of $A$, where $B$ has at least 4 elements. Let $a \in B$ and let $B^{\prime}=B-\{a\}$. Since $B \cap B^{\prime}$ has at least 3 elements, $B^{\prime} \notin C$. Thus, when $B^{\prime}$ replaces $B$ in $C$, the number of subsets in $C$ does not decrease and $C$ still meets the requirements of the problem. Repeating this process, we can always ensure that each subset in $C$ has no more than 3 elements and the number of subsets in $C$ does not decrease. Therefore, the number of subsets in $C$ is no more than 175.
In summary, the maximum value of $k$ is 175.