It may be possible to isolate x, y, z in terms of xyz (say t) as follows:
$x(1+y+yz)=12$ [From the 1st equation]
Let us find y+yz in terms of t from the 2nd equation.
$y+yz=21-xyz=21-t$
Now, this y+yz can be substituted in the 1st step above.
$x(1+21-t)=12$
$\therefore x = \frac{{12}}{{22 - t}}$
Also, $y(1+z+xz)=21$ [From the 2nd equation]
$z+xz=30-t$ [From the 3rd equation]
So, $y(1+30-t)=21$
$\therefore y = \frac{{21}}{{31 - t}}$
Putting these values of x and y in the 1st equation $x+xy+t=12$ or $x(1+y)+t=12$,
$\left( {\frac{{12}}{{22 - t}}} \right)\left( {1 + \frac{{21}}{{31 - t}}} \right) + t = 12$
$ \Rightarrow \left( {\frac{{12}}{{22 - t}}} \right)\left( {\frac{{52 - t}}{{31 - t}}} \right) + t = 12$
$ \Rightarrow 12(52 - t) + t(22 - t)(31 - t) = 12(22 - t)(31 - t)$
$ \Rightarrow 624 - 12t + 682t - 53{t^2} + {t^3} = 8184 - 636t + 12{t^2}$
$ \Rightarrow {t^3} - 65{t^2} + 1306t - 7560 = 0$
t=10 satisfies the above equation.
$ \Rightarrow {t^2}(t - 10) - 55t(t - 10) + 756(t - 10) = 0$
$ \Rightarrow (t - 10)({t^2} - 55t + 756) = 0$
$ \Rightarrow (t - 10)(t - 27)(t - 28) = 0$
t has three possible values: 10, 27, 28.
x and y have already been obtained in terms of t. z can also be found in terms of t in the same fashion.
$ z(1 + 12 - t) = 30$
$ \Rightarrow z = \frac{{30}}{{13 - t}}$
$\left\{ {x,y,z} \right\} \equiv \left\{ {\frac{{12}}{{22 - t}},\frac{{21}}{{31 - t}},\frac{{30}}{{13 - t}}} \right\}$
$ \equiv \left\{ {1,1,10} \right\},\left\{ { - \frac{{12}}{5},\frac{{21}}{4}, - \frac{{15}}{7}} \right\},\left\{ { - 2,7, - 2} \right\}$
