We can solve this with the stars and bars method.
We need to find all non-negative solutions $a+b+c=11$ with $a\leq 8$ and $b,c\leq 9$. (the number is going to be $\overline{(a+1)bc}$).
To do this we first find all the solutions disregarding the upper bound on a,b,c.
This is simple with stars and bars. There are $11$ stars and $2$ bars, so $\binom{13}{2}=78$.
We now need to subtract the solutions in which $a\geq 9,b\geq 10$ and $c\geq 10$.(but none of them happen at the same time, so it is easy)
There are $6$ when $a\geq9$, $3$ when $b\geq 10$ and $3$ when $c\geq 10$.
Therefore the final answer is $78-6-3-3=66$
Let $n=k^3.$ If $n$ has $6$ digits, the largest it could be is $(6\cdot 9)^3 = 157464$, which shows that one digit must be $1$. So in fact the largest it could be is $(1+5\cdot 9)^3 = 97336$ which is too small. So there are no $6$ digit (or larger) solutions. (I think this is the gist of @dezdichado's comment.)
Because the cube root of $100,000 = 46.4\ldots$, we know that $k\leq 46.$ Using your observation that $k\equiv -1, 0, 1 \pmod{9}$ leaves us with only these $16$ possibilities for $k$:
$$1, 8, 9, 10, 17, 18, 19, 26, 27, 28, 35, 36, 37, 44, 45, 46.$$
We can just check these or we can play a bit more:
Since $(5\cdot 9)^3 = 91125$, we see that a $5$-digit solution must have at least one non-$9$ digit. Since $(4\cdot 9+8)^3 = 85184$, the largest the first two digits can be is $79$, so at least one digit is at most $7$. Since $4\cdot 9 +7 = 43$ we can cross off the last three numbers from the list.
Of these, $1, 8, 17, 18, 26, 27$ work.
A couple of observations: It surprises me that the last two pairs are consecutive, and both are of the shape $-1, 0 \pmod{9}$ so I feel like I've missed something. Since $1$ is a solution, it doesn't seem like modular considerations would eliminate $1 \pmod{9}$.
Best Answer
Divide both sides by 3. $$669x-7y=641$$
Observe $669-7\cdot4=641$, so $(x,y)=(1,4)$ is one solution to the equation.
The general solution would be $(x,y)=(1+7k,4+669k)$ where $k$ is an integer. You can check this by substituting it back into the equation.
Since $x>1$ and we want to minimize $2x+3y$, take $k=1$. $$\implies 2x+3y=2(1+7)+3(4+669)=2035$$
Therefore, the answer: $10$.