Finding $a + b + c$ for the $101$st number of the form $3^a + 3^b + 3^c$ when these numbers are listed in increasing order $a,b,c\neq0$ & $a,b,c$ are distinct & $a,b,c\in \mathbb{N}$
numbers in form of
$3^a+3^b+3^c$
If we order them in increasing order, what is the sum of 
$a+b+c=?$
for the $101$st such number?
There are some things I need to clarify about the question a number is represented by $3^a+3^b+3^c$, not $3^a,3^b,3^c$. These do not represent an individual number. 
That being said we could accept $a>b>c$ and then acknowledge that $a,b,c\in \left\{ 0,1,2,3,\cdots,n \right\}$. There are $\dbinom{n+1}{3}$ choices possible for $n$. Therefore, $n$ must give a number greater than $101$. I have chosen $\dbinom{10}{3}=120$, so $n=9$. But I also used $\dbinom{9}{3}=84$ to know which number is $3^9+3^1+3^0$ (it is the $84$th).  I couldn't spot how I would proceed in getting the $101$st number? What do you propose?  

At the top of your question you say $a,b,c\neq 0$ but then the rest of what you say seems to include the possibility that one of them can be $0$. I'll assume that $0$ is not allowed in what follows. Obviously if you include $3^0$ as a possible power the answer will be different, but the method is the same. 
As you say, $\binom 93=84$ and $\binom{10}3=120$. So there are $84$ numbers of the form $3^a+3^b+3^c$ where $a,b,c$ are distinct elements of $\{1,2,...,9\}$, and another $36$ if we allow $10$. All of these are less than $3\times3^{10}=3^{11}$, so these are the first $120$ numbers of the form $3^a+3^b+3^c$ where $a,b,c$ are distinct positive integers.
Any choice of $a,b,c$ where the largest is $10$ will be bigger than any choice where they are all less than $10$, so the $84$ where all powers are at most $9$ come first, and then you want the $17$th number of the form $3^{10}+3^b+3^c$, where $b,c$ are distinct positive integers less than $10$.
Hint for finishing from here: how many choices for $b,c$ are there if $b,c\in\{1,2,...,6\}$?
