can one select 102 17-element subsets of a 102-element set so that the intersection of any two of the subsets has at most 3 elements 
Can one select 102 17-element subsets of a 102-element set so that the intersection of any two of the subsets has at most 3 elements?

I'm not sure how to approach this problem. I think it might be useful to try to find a generalization of the result so one can work with smaller numbers and try to find useful lemmas/properties. In particular, $17$ is prime, so one could replace that with $p$. Then $p(p+1)/3 = 102.$ So it might be reasonable to guess that one can always select $p(p+1)/3$ $p$-element subsets of a $p(p+1)/3$-element set so that the intersection of any two of the subsets has at most $3$ elements. Also, $17\cong 2\mod 3, 17\cong 2\mod 5.$ For $p=2,3$ the problem is straightforward. For $p=5,$ we need to find $10$ $5$-element subsets of a 10-element set so the intersection of any two has at most $3$ elements. I'm not sure how to find the subsets in this case.
It might be useful to consider something related to modular arithmetic modulo the prime p.

This is impossible.
What you are looking at is a constant weight binary code. You can apply the (see Theorem 2 Wikipedia) Johnson bound to get an upper bound on the number of codewords.
Here $q=2,w=17,n=102,$ giving $d=2\times17-3=31$ and $e=(31+1)/2=16.$ I get that the maximum possible number of codewords in such a code must obey
$$
A_2(102,31,17)\leq \left\lfloor \frac{102}{17} 
\left\lfloor \frac{101}{16}  
\right\rfloor \right\rfloor=36
$$
since
$$
A_q(n,d,w)\leq  \left\lfloor \frac{nq^\ast}{w} 
\cdots \left\lfloor \frac{(n-w+e)q^\ast}{e}  
\right\rfloor \cdots \right\rfloor
$$
with $q^\ast=q-1$ since $d\leq 2w.$
