Example 5 Let $S=\{1,2,3, \cdots, 98,99,100\}$, if a three-element subset $A=\{a, b, c\}$ of $S$ satisfies $a+b=3 c$, then $A$ is said to have property $P$. Find the number of all three-element subsets of $S$ that have property $P$.
Let $S_{k}=\{n \mid n \equiv k(\bmod 3), n \in S\}, k=0,1,2$. Then $\left|S_{0}\right|=\left|S_{2}\right|=$ $33,\left|S_{1}\right|=34$.

If $A=\{a, b, c\}$ has property $P$, then either $a, b \in S_{0}$, or one of $a, b$ belongs to $S_{1}$ and the other to $S_{2}$. When $a, b$ both belong to $S_{0}$ and $a \neq b$, the unordered pair $\{a, b\}$ has $C_{33}^{2}$ possibilities. When one of $a, b$ belongs to $S_{1}$ and the other to $S_{2}$, the unordered pair $\{a, b\}$ has $C_{33}^{1} C_{34}^{1}$ possibilities. Therefore, the number of unordered pairs $\{a, b\}$ that satisfy $a+b \equiv 0(\bmod 3)$ is $C_{33}^{2}+C_{33}^{1} C_{34}^{1}$. However, among these, the unordered pairs $\{a, b\}$ where $a=2 b$ or $b=2 a$ do not form a three-element set $A$, as $c=b$ or $c=a$ cannot be achieved. Such unordered pairs are $\{k, 2 k\}(k=1,2, \cdots, 50)$, totaling 50. Therefore, the number of three-element subsets with property $P$ is $C_{33}^{2}+C_{33}^{1} C_{34}^{1}-50=1600$.