How to prove or disprove $(x+y)(x+z)(z+y)=txyz$ has infinitely many solutions.

Question

I thought about asking this type of question after I've seen a question in this site which was the following; $$(x+y)(y+z)(x+z)=13xyz$$ We have shown that this equation has infinitely many solutions with trivial solutions that made us plot $x,y,z=0$ and so forth...

However now I wonder if I can have an approach like this;

$$(x+y)(y+z)(x+z)=txyz$$ $x,y,z,t \in \mathbb{Z} \quad t$ is a constant, no constraints about being positive or negative I can comfortably say that if $x=0$ and $y=0$ there are infinitely many $z$'s If $y,z=0$ there are infinitely many $x$'s so, does that prove $$(x+y)(x+z)(y+z)=txyz$$ for $$x,y,z,t \in \mathbb{Z}$$ has infinitely many solutions?

Answer 1

The equation given above is shown below:

$(x+y)(x+z)(z+y)=t(xyz)$

For $t=8$ there is a numerical solution $(x,y,z)=(1,-6,15)$

Also for $t=13$ there is a parametric solution & is given below:

$x=90k^2-120k+40$

$y=27k^2-36k+12$

$z=135k^2-180k+60$

for $k=0$ we get another solution $(x,y,z)=(10,3,15)$

Answer 2

We have known solution $(x,y,z)=(3,2,10)$ since $(z/y)=5$, take $z=5y$, after substitution & simplifying we get:

$6(x+y)(x+5y)=65xy$

After parametrising above, at $(x,y)=((3+t),(2+kt))$ we get the result.