6.202. $\left\{\begin{array}{l}\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3, \\ \frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x}=3, \\ \frac{1}{x y z}=1 .\end{array}\right.$
## Solution.

Domain of definition: $\left\{\begin{array}{l}x \neq 0, \\ y \neq 0, \\ z \neq 0 .\end{array}\right.$

Multiplying both sides of each equation in the system by $x y z \neq 0$, we get

$\left\{\begin{array}{l}y z+x z+x y=3 x y z, \\ z+x+y=3 x y z, \\ x y z=1\end{array} \Leftrightarrow\left\{\begin{array}{l}y z+x z+x y=3, \\ x+y+z=3, \\ x y z=1,\end{array} \Leftrightarrow\left\{\begin{array}{l}z(y+x)+x y=3, \\ z=3-(x+y), \\ x y z=1 .\end{array} \Rightarrow\right.\right.\right.$

$\Rightarrow\left\{\begin{array}{l}(3-(x+y))(x+y)+x y=3, \\ x y(3-(x+y))=1 .\end{array} \Rightarrow x y=\frac{1}{3-(x+y)} \Rightarrow(3-(x+y))(x+y)+\right.$

$+\frac{1}{3-(x+y)}=3 \Leftrightarrow(x+y)^{3}-6(x+y)^{2}+12(x+y)-8=0$ for $x+y \neq 3$.

Representing the obtained equation as $((x+y)-2)^{3}=0$, from which $x+y=2 ;$ then $z=1, x y=1$.

From this, $\left\{\begin{array}{l}x+y=2, \\ x y=1,\end{array} \Rightarrow y=1, x=1\right.$.

Answer: $(1 ; 1 ; 1)$.