$a+b+c=x+y+z$ and $abc=xyz$ , in which each two of them unequal.

Here is a simple parametrization of a family of solutions.

Let $R,S,T,U$ be any four positive integers such that $RS=TU$.

Then a solution is given by $$\{a,b,c\}=\{R+1,S+1,TU+T+U+1\},\{x,y,z\}=\{T+1,U+1,RS+R+S+1\}.$$

Example

$2\times6=3\times4$ so let $R=2,S=6,T=3,U=4$.

This gives the solution $\{3,7,20\},\{4,5,21\}$.

All solutions can be generated by the following rather more involved procedure. This was obtained by noting that since $a$ is a factor of $xyz$ we can let $a=x_1y_1z_1$ where $x_1$ divides $x$, ... etc.

Choose the numbers $x_2,x_3,y_1,y_3,z_1,z_2$ arbitrarily.

Define $A=y_1z_1-x_2x_3,B=x_2z_2-y_1y_3,C=x_3y_3-z_1z_2$.

Then let $x_1,y_2,z_3$ be any solution of $$Ax_1+By_2+Cz_3=0.$$

The required triples of numbers are then $\{x_1y_1z_1,x_2y_2z_2,x_3y_3z_3\},\{x_1x_2x_3,y_1y_2y_3,z_1z_2z_3\}$.

Example

Let $x_2=x_3=y_1=y_3=1,z_1=2,z_2=4$

Then $A=1,B=3,C=-7$.

The general solution of $x_1+3y_2-7z_3=0$ is $x_1=7r-3s,y_2=s,z_3=r$.

The required triples of numbers are then $\{r,4s,14r-6s\},\{8r,s,7r-3s\}$ for 'any' choice of $r$ and $s$.