If $x + y = 5xy$ , $y + z = 6yz$ , $z + x = 7zx$ . Find $x + y + z$ . 
If $x + y = 5xy$ , $y + z = 6yz$ , $z + x = 7zx$ . Find $x + y + z$ .

What I Tried :
I used some clever ways to get $x + y + z = 26xyz$ , but I suppose we have some solution as a number .
All all $3$ to get :-
$$2(x + y + z) = 5xy + 6yz + 7zx$$
Or,
$$ 2(x + y + z) = (xy + xy + xy + xy + xy) + (yz + yz + yz + yz + yz + yz) + (zx + zx + zx + zx + zx + zx + zx)$$
That is,
$$ 2(x + y + z) = (xy + zx) + (xy + zx) + (xy + zx) + (yz + zx) + (yz + zx) + (yz + zx) + (yz + zx) + (xy + yz) + (xy + yz)$$
Now see that $(xy + zx) = x(y + z) = 6xyz$ , similarly $(yz + zx) = 5xyz$ and $(xy + yz) = 7xyz$
So
$$2(x + y + z) = 3(6xyz) + 4(5xyz) + 2(7xyz)$$
$$\Rightarrow (x + y + z) = \frac{52xyz}{2} = 26xyz$$
I tried till this , then I have no idea . Can anyone help?

Assuming $x,y,z \neq 0$, we have that,
$$x + y = 5xy \iff \frac1x+\frac1y =5$$
$$y + z = 6yz \iff \frac1y+\frac1z =6$$
$$z + x = 7zx \iff \frac1z+\frac1x =7$$
then solve for $1/x$, $1/y$, $1/z$.
The case $x=0 \lor y=0 \lor z=0$ is trivial.
