Without loss of generality assume $ a+b+c\ge x+y+z$ (otherwise just switch $ (a,b,c)$ with $ (x,y,z)$), then $ a+b+c\ge abc$. We then have two cases: 1. $ c=1$. Then, $ a+b+c\ge abc\Leftrightarrow 1+a+b\ge bc\Leftrightarrow 2\ge ab-a-b+1$ $ \Leftrightarrow 2\ge (a-1)(b-1)$. Then, because $ a\ge b$, we have either $ b=1$, or $ b=2$ then $ a=2$ or $ a=3$. Therefore, the only triplets $ (a,b,c)$ satisfying $ c=1$ and $ a+b+c\ge abc$ are $ (a,1,1), (2,2,1), (3,2,1)$. 2. $ c\ge 2$. Then, $ a+b+c\ge abc\rightarrow a+b+c\ge 2ab\Leftrightarrow 1\ge ab-a-b+1+ab-c\leftrightarrow 1\ge (a-1)(b-1)+(ab-c)$ (*) However, as $ a\ge b\ge c\ge 2$ then $ (a-1)(b-1)\ge 1$ and $ ab-c>1$ so (*) is false, so the case when $ c\ge 2$ fails. Now, if $ (a,b,c)=(2,2,1)$ have $ x+y+z=4, xyz=5$ but then one of $ x,y,z$ is equal to $ 5$ and as $ x,y,z\ge 1$ we have $ x+y+z>4$ so we cannot have $ (a,b,c)=(2,2,1)$. If $ (a,b,c)=(3,2,1)$ have $ x+y+z=6,xyz=6$ then $ x,y,z\in \{1,2,3,6\}$; none of $ x,y,z$ is $ 6$ as otherwise $ x+y+z>6$. Then one of $ x,y,z$ is $ 3$; as $ x\ge y\ge z$, $ x=3$ then $ y+z=3,yz=2$ so $ y=2,z=1$. Hence get $ (x,y,z)=(3,2,1)$. If $ (a,b,c)=(a,1,1)$ have $ x+y+z=a$ and $ xyz=a+2$. Because $ x\ge y\ge z$ and $ x+y+z=a$ we must have $ a\ge \frac{a}{3}$ then, if $ yz\ge 5$ then $ x\le \frac{a+2}{5}$ so $ \frac{a}{3}\le\frac{a+2}{5}$ so $ a\le 3$ which is impossible as $ a+2=xyz\ge 5x>5\rightarrow a>3$. Therefore, $ yz\le 4$ If $ yz=4$, we can have $ y=z=2$ giving $ x+4=a, 4x=a+2$ so $ x=2,a=6$ and get $ (a,b,c)=(6,1,1), (x,y,z)=(2,2,2)$, or we can have $ z=1,y=4$ then $ x+5=a,4x=a+2\rightarrow 3x=7$ which does not give an integer value for $ x$, so $ (y,z)=(4,1)$ is impossible. If $ yz=3$, $ y=3,z=1$ then $ x+4=a, 3x=a+2$ so $ x=3,a=7$ and get $ (a,b,c)=(7,1,1), (x,y,z)=(3,3,1)$. If $ yz=2$, $ y=2,z=1$ then $ x+3=a,2x=a+2$ so $ x=5,a=8$ so get $ (a,b,c)=(8,1,1), (x,y,z)=(5,2,1)$. If $ yz=1$, $ y=z=1$ then $ x+2=a, x=a+2$ which is impossible. Therefore, if $ a+b+c\ge x+y+z$ we get the following six-tuplets $ (a,b,c,x,y,z)$: $ (3,2,1,3,2,1);(6,1,1,2,2,2);(7,1,1,3,3,1);(8,1,1,5,2,1)$.If $ x=y+z\ge a+b+c$ by symmetry of the system and teh conditions on $ a,b,c,x,y,z$ we just switch $ (a,b,c)$ with $ (x,y,z)$. Therefore, all the six-tuplets satisfying the conditions of the problem are: $ (3,2,1,3,2,1);(6,1,1,2,2,2);(7,1,1,3,3,1);(8,1,1,5,2,1), (2,2,2,6,1,1);(3,3,1,7,1,1);(5,2,1,8,1,1)$

Consider $ (ab-1) (c-1)$ , $ (a-1)(b-1)$, $ (xy-1)(z-1)$, $ (x-1)(y-1)$. When you add up these 4 numbers, we get that $ (ab-1) (c-1) + (a-1)(b-1) + (xy-1)(z-1) + (x-1)(y-1) = 4$. Their non negativity is assured since we are given that \[ a+b+c=xyz,\; x+y+z=abc,\; a\ge b\ge c\ge 1,\; x\ge y\ge z\ge 1.\]. So we just have to check for I think $ \binom {4+4-1}{4-1} = \binom {7}{3} = 35$. We can eliminate a lot of cases though using parity and primes condition etc. Then we get the 7 6tuplet solutions.