One, (20 points) Let $x, y$ be non-negative integers, $x+2y$ is a multiple of 5, $x+y$ is a multiple of 3, and $2x+y \geqslant 99$. Try to find the minimum value of $S=7x+5y$.
Let $x+2 y=5 A, x+y=3 B, A, B$ be integers.
Since $x, y \geqslant 0$, it follows that $A, B \geqslant 0$.
It is also easy to see that $x=6 B-5 A, y=5 A-3 B$, thus,
$$
\begin{array}{l}
2 x+y=9 B-5 A \geqslant 99, \\
S=7 x+5 y=27 B-10 A .
\end{array}
$$

Therefore, the problem becomes: For integers $A, B \geqslant 0, 6 B \geqslant 5 A \geqslant 3 B, 9 B \geqslant 5 A+99$, find the minimum value of $S$.

Since $15 A \geqslant 9 B \geqslant 99+5 A$, it follows that $10 A \geqslant 99, A \geqslant 10$, then $9 B \geqslant 50+99=149, B \geqslant 17$.
Thus, $5 A \geqslant 51, A \geqslant 11$.
Furthermore, $9 B \geqslant 55+99=154, B \geqslant 18$.
If $B \geqslant 19$, then
$$
S=9 B+2(9 B-5 A) \geqslant 9 \times 19+2 \times 99=369.
$$

If $B=18$, then $5 A \leqslant 9 B-99=63, A \leqslant 12, A$ can only be 12 or 11. Among these, $A=12$ makes $S=27 \times 18-10 \times 12=366$ the minimum value, and this is the only case (at this point, $x=48, y=6$).