I am trying to solve the following type of general 3-variable diophantine equation: $$axyz + bxy + cxz + dyz + ex + fy +gz + h=0$$
(The variables are $x$, $y$ and $z$).
I know how to solve the less complicated 2-variable variant: $$ axy + bx + cy + d =0 $$ After multiplying by $a$, the equation becomes $a^2xy + abx + acy + ad = 0$. From that point, we can complete the square and write $(ax+c)(ay+b)=bc-ad$.
The term on the right-hand side can be factored, and then matched against the left-hand side to obtain solutions.
I can't really see how to elegantly generalize this to the 3-variable case.
I tried multiplying by $a^2$ and factoring into $(ax+d)(ay+c)(az+b)$, but the latter's expansion gives $a^3xyz + a^2bxy + a^2cxz + a^2dyz + abcx + abdy + acdz + bcd$, while the original equation multiplied by $a^2$ is $a^3xyz + a^2bxy + a^2cxz + a^2dyz + a^2ex + a^2fy + a^2gz + a^2h = 0$.
Doing it that way looks like it might not work if $a^2e \ne abc$ for example.
Is there any other way to solve it?