Logarithms!

We can rewrite the equations as following

$\log _{ 2 }{ x } +\log _{ 4 }{ y } +\log _{ 4 }{ z } =\log _{ 4 }{ { x }^{ 2 }yz } =2\Rightarrow { x }^{ 2 }yz={ 2 }^{ 4 }$

$\log _{ 3 }{ y } +\log _{ 9 }{ z } +\log _{ 9 }{ x } =\log _{ 9 }{ { y }^{ 2 }zx } =2\Rightarrow { y }^{ 2 }zx={ 3 }^{ 4 }$

$\log _{ 4 }{ z } +\log _{ 16 }{ x } +\log _{ 16 }{ y } =\log _{ 16 }{ { z }^{ 2 }xy } =2\Rightarrow { z }^{ 2 }xy={ 4 }^{ 4 }$

We multiply the solutions together to get that ${ \left( xyz \right) }^{ 4 }={ 24 }^{ 4 }\Rightarrow xyz=24$

From this it's easy to get the values of $x,y,z$ which are equal to $\frac { 2 }{ 3 } ,\frac { 27 }{ 8 } ,\frac { 32 }{ 3 }$ respectively.

Therefore the answer is $2+3+27+8+32+3=75$ .

The given equations are

The first equation can be written as

$\log_{2}{x}+\log_{4}{y}+\log_{4}{z}=\log_{4}{x^2yx}=2 \Rightarrow {x^2}yz=2^4............\boxed{1}$

Similarly,

$\log_{3}{y}+\log_{9}{z}+\log_{9}{x}=2 =\log_{9}{{y^2}xz} \Rightarrow {y^2}zx = 3^4............\boxed{2}$

Similarly,

$\log_{4}{z}+\log_{16}{x}+\log_{16}{y}=2=\log_{16}{{z^2}xy} \Rightarrow {z^2}xy=16^2=4^4...\boxed{3}$

Multiplying $1,2,3$ ,we get,

$(xyz)^{4}=2^4 \times 3^4 \times 4^4 = (24)^4 \Rightarrow \boxed{xyz=24}$

Then $1$ becomes ${x^2}yz=16 \Rightarrow x(xyz)=16 \Rightarrow x=\dfrac{16}{24} \Rightarrow \boxed{x=\dfrac{2}{3}}$

Similarly $2$ becomes ${y^2}xz=81 \Rightarrow y(xyz)=81 \Rightarrow y=\dfrac{81}{24} \Rightarrow \boxed{y=\dfrac{27}{8}}$

Similarly $3$ becomes ${z^2}xy = 256 \Rightarrow z(xyz)=256 \Rightarrow z=\dfrac{256}{24} \Rightarrow \boxed{z=\dfrac{32}{3}}$

Therefore the answer is $2+3+27+8+32+3=\boxed{75.}$