Homogeneous diophantine equation $x^3+2y^3+6xyz=3z^3$ Is it known if there are infinitely (non-proportional) many integer solutions
to $x^3+2y^3+6xyz=3z^3$ ? 
Motivation : if true, this would provide an alternative solution to that recent
MSE question, by putting $a=\frac{3z^2}{xy},b=-\frac{x^2}{yz},c=-\frac{2y^2}{xz}$.
 A: It seems to be possible to use the fact that $x^3+2y^3+6xyz = 3z^3$ is an elliptic curve with Weierstrass form $y^2 + 6 x y + 16 y = x^{3} - 105 x^{2}$ and that it has rank $1$ to produce integer solutions. I used the following Sage script to do so experimentally.
R.<x, y, z> = QQ[]
eq = x^3+2*y^3+6*x*y*z-3*z^3
P = [1,1,-1]

E = EllipticCurve_from_cubic(eq, P, morphism=False)
f = EllipticCurve_from_cubic(eq, P, morphism=True)

G = E.gens()[0]
for n in [2..10]: # adjust for more solutions
    P = f.inverse()(n*G)
    a = P[0]
    k = a.denominator()
    a = a * k
    b = P[1] * k
    c = P[2] * k
    if a in ZZ and b in ZZ and c in ZZ:
        print a,b,c, a^3+2*b^3+6*a*b*c-3*c^3

I am not sure if it can be proven or not, that this will produce infinitely many non-proportional integer solutions, but atleast it seems to do so experimentally.
A: It's a homogeneous equation, so if $(x,y,z)$ is a solution then $(kx,ky,kz)$ is also a solution:
\begin{eqnarray*}
(kx)^3+2(ky)^3-3(kz)^3 &=& 0 \\ \\
k^3x^3 + 2k^3y^3 - 3k^3z^3 &=& 0 \\ \\
k^3(x^3 + 2y^3-3y^3) &=& 0
\end{eqnarray*}
If it has one, non-zero solution then it will have infinitely many solutions.
Clearly $(x,y,z)=(1,1,1)$ is a solution, and so $(x,y,z)=(k,k,k)$ are all solutions.
