Let $(x_1,y_1,z_1)and(x_2,y_2,z_2)$  where $(x_1>x_2)$  be two triplets satisfying the following simultaneous equations:
 ${\log }_{10}\left(2xy\right)=\left({\log }_{10}x\right)\left({\log }_{10}y\right)$
 ${\log }_{10}\left(yz\right)=\left({\log }_{10}y\right)\left({\log }_{10}z\right)$
 ${\log }_{10}\left(2zx\right)=\left({\log }_{10}z\right).\left({\log }_{10}x\right)$
Then the value of ${(x_1+y_1+z_1)}^{x_2{y}_2{z}_2}$ is:
 

Let  ${\log }_{10}x=a,{\log }_{10}y=b,{\log }_{10}z=c$
Hence, given equation are $a+b+{\log }_{10}2=ab$ ----------------(1)
      $b+c=bc$  ----------------(2)
           $c+a+{\log }_{10}2=ca$  ----------------(3)
Now, (1) – (3)
 $\Rightarrow b-c=a(b-c)$   $\Rightarrow b=c$ or  $a=1$.
Putting b=c in equation (2), we get                                                                       

 or  $b=2$
Putting this in equation(1),  $b=0\Rightarrow a+{\log }_{10}2=0\Rightarrow {\log }_{10}2x=0\Rightarrow x=1/2$             $2b=b^2\Rightarrow b=0$
             $b=2\Rightarrow a+2+o{g}_{10}2=2a$  $a=lo{g}_{10}200\Rightarrow x=200$
Now, a=1 is rejected, as by putting this in first equation. 
$1+b+{\log }_{10}2=b\Rightarrow 1+{\log }_{10}2=0$ which is not possible.
       $(x_1,y_1,z_1)=(200,100,100)$            $(x_2,y_2,z_2)=(\frac{1}{2},1,1)$  
         $\therefore {(x_1+y_1+z_1)}^{x_2{y}_2{z}_2}={\left(400\right)}^{1/2}=20$  .