Around the diophantine equation $\frac{a}{2b+3c}+\frac{b}{2c+3a}+\frac{c}{2a+3b}=\text{odd integer}$, over positive integers I am interested to know if a similar theorem that shows this answer of the post 
Estimating the size of solutions of a diophantine equation (this MathOverflow, January 5th 2016) is feasible for a different diophantine problem.
Conjecture. Let $a,b$ and $c$ be integers greater or equal than $1$. Then $$\frac{a}{2b+3c}+\frac{b}{2c+3a}+\frac{c}{2a+3b}$$ can never be an odd (positive) integer.

Question. I would like to know if the reasonings in the answers of the linked post also works for the diophantine equation in previous conjecture. Is it possible to prove previous conjecture, or is there any counterexample? Many thanks.

I don't know if this equation is in the literature (as reference I've added the mentioned post and [1]) as a special case of the expression $$\frac{\lambda a}{\beta b+\eta c}+\frac{\lambda b}{\beta c+\eta a}+\frac{\lambda c}{\beta a+\eta b}=N$$
with $N$ a non-zero integer, and for  given integers $1\leq \lambda\leq \beta\leq \eta$. Our case, is similar than the linked post with $(\lambda,\beta,\eta)=(1,2,3)$. Our curve can be written (I did the calculation using Wolfram Alpha online calculator) 
$$6 a^3 + 9 a^2 b + 4 a b^2 + 6 b^3 + 4 a^2 c + 18 a b c + 9 b^2 c + 9 a c^2 + 4 b c^2 + 6 c^3$$
$$\qquad\qquad\qquad\qquad\qquad\qquad\qquad-n((2 a + 3 b) (3 a + 2 c) (2 b + 3 c))=0$$
from $\frac{a}{2b+3c}+\frac{b}{2c+3a}+\frac{c}{2a+3b}=n$.
I think that this is a good companion of previous post, but if it is in the literature feel free to comment or provide your answer as a reference request and I search and read the statement about previous Conjecture from the literature.
References:
[1] Andrew Bremner and Allan MacLeod, An Unusual Cubic Representation Problem, Annales Mathematicae et Informaticae Volume 43 (2014), pp. 29-41.
 A: The problem of this question is qualitatively and quantitatively different in some ways from that considered by Andrew Bremner and myself.
If we take the cubic, with $N$ as a fixed constant, it is possible to show that the related elliptic curve is
\begin{equation*}
E_N:G^2=H^3+((35N+18)H+4(1260N+1441))^2
\end{equation*}
with the formulae linking $(H,G)$ to solutions $a,b,c$ being lengthy, but straightforward to find.
The discriminant is
\begin{equation*}
\Delta=2^{12}5^27^2(5N-3)(1260N+1441)^3(7N^2+15N+9)
\end{equation*}
so that $E_N$ has two components for all $N\ge1$.
This curve is in the standard form for one with $\mathbb{Z}3$ torsion, with points of order $3$ when $H=0$.
The algebra reducing the original cubic to the elliptic curve showed that there are other rational points.
I found
\begin{equation*}
H=\frac{288(18N+31)(36N+43)}{361}
\end{equation*}
giving
\begin{equation*}
G=\pm \frac{4(468N+635)(79056N^2+186120N+115297)}{6859}
\end{equation*}
which just lead to trivial solutions.
What I then found was the very surprising fact that this point is double $(-144,\pm 2660)$ for all $N$.
The positive $G$ value gives the parametric solution
\begin{equation*}
a=12(34992N^2+78120N+40499)
\end{equation*}
\begin{equation*}
b=-279936N^2-1271340N-1096057
\end{equation*}
\begin{equation*}
c=18(10368N^2+54180N+52811)
\end{equation*}
The elliptic curve has positive rank and simple numerical experiments show rank $3$ is fairly common.
