smc

We apply casework to this problem. The only sets that contain two multiples of seven are those for which: 1. The multiples of $7$ are $1(\mathrm{mod}10)$ and $8(\mathrm{mod}10).$ That is, the first and eighth elements of such sets are multiples of $7.$ The first element is $1+10k$ for some integer $0\le k\le 99.$ It is a multiple of $7$ when $k=2,9,16,\dots ,93.$ 3. The multiples of $7$ are $2(\mathrm{mod}10)$ and $9(\mathrm{mod}10).$ That is, the second and ninth elements of such sets are multiples of $7.$ The second element is $2+10k$ for some integer $0\le k\le 99.$ It is a multiple of $7$ when $k=4,11,18,\dots ,95.$ 5. The multiples of $7$ are $3(\mathrm{mod}10)$ and $0(\mathrm{mod}10).$ That is, the third and tenth elements of such sets are multiples of $7.$ The third element is $3+10k$ for some integer $0\le k\le 99.$ It is a multiple of $7$ when $k=6,13,20,\dots ,97.$ Each case has $\lfloor \frac{100}{7}\rfloor =14$ sets. Therefore, the answer is $14\cdot 3=\boxed{42}.$