Solution:

We will show by induction that for all $n \geq 3$, it holds that at most $n-2$ powers of three are sums of three distinct elements of $S$ for any set $S$ of positive real numbers with $|S|=n$. This is trivially true when $n=3$.

Let $n \geq 4$ and consider the largest element $x \in S$. The sum of $x$ and any two other elements of $S$ is strictly between $x$ and $3x$. Therefore $x$ can be used as a summand for at most one power of three. By the induction hypothesis, at most $n-3$ powers of three are sums of three distinct elements of $S \setminus \{x\}$. This completes the induction.

Even if it was not asked to prove, we will now show that the optimal answer $n-2$ is reached. Observe that the set $S=\{1,2,3^{2}-3,3^{3}-3, \ldots, 3^{n}-3\}$ is such that $3^{2}, 3^{3}, \ldots, 3^{n}$ can be expressed as sums of three distinct elements of $S$. This makes use of the fact that each term of the form $3^{k}-3$ can be used in exactly one sum of three terms equal to $3^{k}$.