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:
$\mathcal{l}o{g}_{10}\left(2xy\right)=\left(\mathcal{l}o{g}_{10}x\right)\left(\mathcal{l}o{g}_{10}y\right), \mathcal{l}o{g}_{10}\left(yz\right)=\left(\mathcal{l}o{g}_{10}y\right)\left(\mathcal{l}o{g}_{10}z\right)$  and  $\mathcal{l}o{g}_{10}\left(2zx\right)=\left(\mathcal{l}o{g}_{10}z\right).\left(\mathcal{l}o{g}_{10}x\right)$
Then the value of  ${(x_1+y_1+z_1)}^{x_2{y}_2{z}_2}$  is

Let  $\mathcal{l}o{g}_{10}x=a,\mathcal{l}o{g}_{10}y=b,\mathcal{l}o{g}_{10}z=c$
Hence, given equation are  $a+b+\mathcal{l}o{g}_{10}2=ab$ ……… (1)
  $b+c=bc$  ……… (2)
  $c+a+\mathcal{l}o{g}_{10}2=ca$  ……… (3)
Now, (1) – (3)
$\Rightarrow b-c=a(b-c)$   $\Rightarrow b=c$  or a = 1.
Putting b = c in equation (2), we get  $\therefore 2b=b^2\Rightarrow b=0$
Putting this in equation (1),  $b=0\Rightarrow a+\mathcal{l}o{g}_{10}2=0\Rightarrow \mathcal{l}o{g}_{10}2x=0\Rightarrow x=1/2$
 $b=2\Rightarrow a+2+\mathcal{l}o{g}_{10}2=21$      $\therefore a=\mathcal{l}o{g}_{10}200\Rightarrow x=200$  
Now, a = 1 is rejected, as by putting this in first equation.
 $1+b+\mathcal{l}o{g}_{10}2=b\Rightarrow 1+\mathcal{l}o{g}_{10}2=0$ which is not possible.
$\therefore (x_1,y_1,z_1)=(200,100,100)$           $(x_2,y_2,z_2)=(\frac{1}{2},1,1)$

$\therefore {(x_1+y_1+z_1)}^{x_2{y}_2{z}_2}={\left(400\right)}^{1/2}=20$