Status of $x^3+y^3+z^3=6xyz$ In

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*Erik Dofs, Solutions of $x^3 + y^3 + z^3 = nxyz$, Acta Arithmetica 73 (1995) pp. 201–213, doi:10.4064/aa-73-3-201-213, EuDML
the author has studied the Diophantine equation
\begin{equation}
x^3+y^3+z^3=nxyz\tag{1}
\end{equation}
where $n$ is an integer. I am interested in the specific case $n=6$. The author mentions:

For a fixed $n$-value, (1) can be transformed into an elliptic curve
with a recursive solution structure derived by the "chord and tangent
process".

How does this work for the case $n=6$? Does this generate all or infinitely many solutions? What progress has been made in this specific case?
 A: My paper
"Unsolvable cases of $P^3+Q^3+cR^3=d PQR$", Rocky Mountain J. of Math. (28),No 3, 1998
gives 7 (infinite) generic classes of unsolvability of the original equation. However
there are still many (generic) cases to prove!
Solutions have been found for all $n$ in the range $0<n\le1000$.
A: (Collecting comments into a community wiki answer.)
There is a standard method for transforming a smooth cubic into Weierstrass form. See for example Section 1.3 or Appendix B of Silverman and Tate's book Rational Points on Elliptic Curves.  It is also implemented in Sage as $\mathtt{EllipticCurve\_from\_cubic}$.
For $n=6$ your curve is isomorphic to $Y^2 + Y = X^3 -54X  - 88$. According to the LMFDB, it has torsion $\mathbb{Z}/3\mathbb{Z}$ and rank 1.  Thus one can start with the obvious solution $(x:y:z) = (1:2:3)$ and its permutations and generate all rational solutions.  Since your equation is homogeneous, finding all rational solutions is equivalent to finding all integer solutions.
The positive solutions constitute one of two connected components of the real locus of the elliptic curve. Since the generator is on that component (and the identity is not), its odd multiples again yield positive solutions. The next batches of positive solutions are permutations of $(1817,3258,5275)$ and $(4904676969, 10840875082, 15051171563)$.
A: @Haran enquired about case for $n=6$.
Well lately Seiji Tomita has given a numerical solution for it.
He used the equation below to arrive at a numerical solution:
$$
(m^2+m+1)^3+(m^2-m+1)^3+(2)^3=(m^2+5)(m^2+m+1)(2)(m^2-m+1).
$$
To arrive at $n=6$ we substitute $m=1$. The downside of this is that, when using the above equation one can get only one numerical solution for $n=6$. The equation gives us:
$$
3^3+1^3+2^3=6(3\cdot2\cdot1).
$$
Tomita has given a general method through which solutions for different $n$ can be arrived at. His "Computational number theory" web page discusses this in #456. $x^3 + y^3 + z^3 = n x y z$.
