Positive integers satisfying $(x+y+z)(xy+yz+zx)=12xyz$ 
Find all positive integers $x$, $y$ and $z$ satisfying $(x+y+z)(xy+yz+zx)=12xyz$.

Firstly, I think of $\textsf{Vieta Jumping}$, but the quadratic coefficient is not a number, so I abandon this thought.
Then I wanted to do something with this homogeneous condition, so I divide the equation by $x$, then we have to find $b\overset{\rm WLOG}\ge c\in\Bbb Q_+$, satisfying
$$(1+b+c)(b+c+bc)=12bc.$$
It seems tempting to do find some algebraic properties. By Cauchy inequality,
$$12bc\ge\left(\sqrt{bc}+\sqrt b+\sqrt c\right)^2.$$
So $12\ge\left(1+\sqrt{\dfrac bc}+\sqrt{\dfrac cb}\right)^2$, then we can solve out the range of $\dfrac bc$.
But none of these is making any fundamental progress. How to solve this?
 A: There are no positive integer solutions for the equation
$$(x+y+z)(xy + yz + zx) = 12xyz$$
The way I do it might be an overkill. I transform the problem to one finding rational points on an elliptic curve. It turns out the elliptic curve has rank zero, so it has finitely many rational points. One can verify none of them give us a positive integer solution.

Since $$(x+y+z)(xy+yz+xz) = (x+y)(y+z)(z+x) + xyz$$
the question at hand is equivalent to finding solutions of
$$(x+y)(y+z)(z+x) = 11xyz$$
for $x,y,z \in \mathbb{Z}_{+}$. Since this equation is homogeneous, it is equivalent to finding solutions for $x,y,z \in \mathbb{Q}_{+}$. This is because any rational solution can be converted to an integer one by clearing the denominators.
We will "normalized" the equation by assuming $x + y + z = 1$. Under this assumption, the equation becomes
$$\left(\frac1x - 1\right)\left(\frac1y - 1\right)\left(\frac1z - 1\right) = 11$$
Call the $3$ factors on LHS as $u,v,w$, we have $(x,y,z) = \left(\frac{1}{1+u},\frac{1}{1+v},\frac{1}{1+w}\right)$ and the problem is equivalent to finding solutions for
$$uvw = 11\quad\text{ and }\quad \frac1{1+u} + \frac{1}{1+v} + \frac{1}{1+w} = 1$$
for $u,v,w\in \mathbb{Q}_{+}$.
Set $u = \frac{11}{vw}$, substitute that in $2^{nd}$ equation and rearrange, we obtain:
$$vw(9-(v+w)) = 11$$
Let $p = v+w$ and $q = v-w$, this becomes
$$(p^2 - q^2)(9 - p) = 44
\iff q^2(9-p) = p^2(9-p) - 44$$
Multiply both sides by $\frac{(-44)^2}{(9-p)^3}$, this leads
to
$$\left(-\frac{44q}{9-p}\right)^2 =
\left(-\frac{44}{9-p}\right)^3 + \left(\frac{44p}{9-p}\right)^2$$
Change variable to $(X,Y) = \left( -\frac{44}{9-p}, \frac{-44q}{9-p} \right) \iff (p,q) = \left( 9 + \frac{44}{X}, \frac{Y}{X} \right)$, the problem transforms to finding rational solution on elliptic curve
$$Y^2 = X^3 + (9X+44)^2$$
Throwing following commands to the online magma calculator,
Q<x> := PolynomialRing(Rationals());  
E00:=EllipticCurve(x^3+(9*x+44)^2);
MordellWeilShaInformation(E00);  
P := Generators(E00)[1];
for i := 1 to 5 do
    print i, i*P;
end for;

one find this elliptic curve has rank $0$ and Torsion subgroup $\mathbb{Z}/6\mathbb{Z}$ consists of 6 elements:
$$(X,Y) = (44,-528),(0,-44),(-4,0),(0,44),(44,528),(\infty,\infty)$$
None of these give us a solution with all $u,v,w > 0$, so the original equation doesn't have any positive rational and hence positive  integer solutions.
A: The given equation can be rewritten as:
$x^2 y+x y^2+x^2 z-9 x y z+y^2 z+x z^2+y z^2 = 0$
The biggest problem of the task is that this is a volume and volumes of that kind are not visibly nice to humans. That has impact an the solutions methodologies. Most of them attempt to give $z(x,y)$ or even make the solution have only two free parameters.
I interpret the task that like the most trivial solution $(0,0,0)$ seeks for triples only. Lacking a constant makes this type of equation not homogenous.



My pictures show a present day solution from a bolide symbolic mathematics solver of the task. This is more that 2 and quarter of pages long and just the solution not the solutions paths.
As could have been expected this heavily shows the $9 x y z$ term. The root symbol is simply square rooting, the Root function determines sets of roots. There is a case for $x$ being less or equal $-1$,$0$ or bigger equal $1$ by the solutions method start with $x$.
Sorry for such small picture size. The layout enforcements of math.stackexchange.com make my pictures smaller than they really are. Click on the picture or download them to properly ready the details. Even better get yourself access to wolframcloud.com and produce the solution yourself for verification purposes. This is thereby the 13.2 version of Mathematica calculating the result.
Setting $x=$ gives pretty easy looking solutions worth for a closer look with simpler methods.
Reduce[(x+y+ z)(x y+y z+ z x)==12x y z,{x,y,z},Integers]
on wolframalpha.com gives some more visual result.

