Show there exists at most $n-2$ powers of $3$ that can be written as the sum of three distinct elements of a set Recently, I have found this problem:

Let $\mathcal{H}$ be a set of $n\geq3$ distinct real numbers. Show that there are at most $n-2$ distinct integers that are a power of $3$ and can be written as the sum of three distinct elements in the set $\mathcal{H}$.

I think that this problem can be done by induction. Namely, I can simply show that when $\mathcal{H}$ has three elements I can construct only one integer power of $3$ combining them.
But, how can we proceed in the demonstration?
 A: Let $p$ be the number of integer powers of 3 in $\mathcal{H}$. Each of the powers of $3$ can be written as $a_i+b_j+c_k$. For example if $\mathcal{H}$ contains $(27,81,243,729)$ then $a_1+b_1+c_1=27,\ a_2+b_2+c_2=81,\ a_3+b_3+c_3=243,\ a_4+b_4+c_4=729$.
In order to minimize $n$ (such that $p$ is as close as possible to $n$), each of the triples $a_i,b_j,c_k$ can overlap with another triple by up to $2$ addends giving something like: $$a_1+b_1+c_1=27,\ b_1+c_1+c_2=81,\ c_1+c_2+c_3=243,\ c_2+c_3+c_4=729$$ or $$a_1 + (-a_1)+27=27,\ a_1 + (-a_1)+81=81,$$$$a_1 + (-a_1)+243=243,\ a_1 + (-a_1)+729=729$$  That means there area minimum of $3+p-1=p+2$ addends across all powers of $3$.
Therefore there are nominally a minimum of $2p+2$ distinct integers needed in $\mathcal{H}$ for $p$ powers of 3. However in the second example above the overlap between powers of 3 and addends reduces that to $p+2$.
Therefore the minimum $n$ for a given $p$ is $n=p+2 \implies n-2=p$
EDIT
To disprove the assertion (at most $n-2$ powers of 3 are sums of 3 distinct elements in $\mathcal{H}$) for a given $n$, $p$ must be maximized. The following shows that when $p$ is maximized it is at most $n-2$, proving the assertion.
First note that the "first" element of the set of powers of 3 (call that subset of  $\mathcal{H}$: $\mathcal{T}$) requires exactly 3 unique addends. Also required is an element from  $\mathcal{T}$. However that element from $\mathcal{T}$ can also be an addend.
Therefore the "first" element of $\mathcal{T}$ accounts for 1 element in  $\mathcal{T}$ and minimally accounts for 3 elements in  $\mathcal{H}$
The rest of elements in $\mathcal{T}$ can optimally have their sum created by selecting two addends that were used to create the sum of the "first" element of  $\mathcal{T}$. The third addend would come from  $\mathcal{T}$.
Therefore each "non-first" elements of $\mathcal{T}$ accountd for 1 additional addend from  $\mathcal{T}$ (which also is the sum) and minimally accounts for 1 element in  $\mathcal{H}$
Adding up the minimum number of elements from $\mathcal{H}$ for the addends and the powers of 3 that the addends sum to gives $p+2$. Therefore $n >= p+2 \implies n - 2 >= p$
