3 Let $X=\{0,1,2, \cdots, 9\}, F=\left\{A_{1}, A_{2}, \cdots, A_{k}\right\}$ where each element $A_{i}$ is a non-empty subset of $X$, and for any $1 \leqslant i<j \leqslant k$, we have $\left|A_{i} \cap A_{j}\right| \leqslant 2$, find the maximum value of $k$. (No.
3. Let $F$ satisfy the conditions. To maximize $|F|$, note that $\left|A_{i} \cap A_{j}\right| \leqslant 2$. We should include some sets with fewer elements in $F$. Clearly, all sets satisfying $\left|A_{i}\right| \leqslant 2$ can be included in $F$. Next, all sets satisfying $\left|A_{i}\right| \leqslant 3$ can also be included in $F$. In fact, consider any two sets $A_{i}$ and $A_{j}$ in $F$. When one of $A_{i}$ or $A_{j}$, say $A_{i}$, satisfies $\left|A_{i}\right| \leqslant 2$, then $\left|A_{i} \cap A_{j}\right| \leqslant 2$; when $\left|A_{i}\right|=\left|A_{j}\right|=3$, if $\left|A_{i} \cap A_{j}\right|>2$, then $\left|A_{i} \cap A_{j}\right|=3$, so $A_{i}=A_{j}$, which is a contradiction. Therefore, $|F|_{\text {max }} \geqslant \mathrm{C}_{10}^{1}+\mathrm{C}_{10}^{2}+\mathrm{C}_{10}^{3}=175$. The $F$ constructed above is saturated, meaning no more sets can be added. However, this does not necessarily mean that $|F|$ is maximized. We now prove that $|F| \leqslant 175$. Suppose $|F|>175$, then there must be a set in $F$, say $A$, such that $|A|>3$. Take any three-element subset $A^{\prime}$ of $A$. Since $\left|A \cap A^{\prime}\right|=3>2$ and $A$ is in $F$, $A^{\prime}$ is not in $F$. Replace $A$ in $F$ with $A^{\prime}$ to get $F^{\prime}$, which also satisfies the conditions. Continuing this process, we can replace all sets in $F$ with more than 3 elements with 3-element sets, resulting in $F^{*}$, which still satisfies the conditions. But at this point, $|F|=\left|F^{*}\right| \leqslant 175$, which is a contradiction.