6.212. $\left\{\begin{array}{l}\frac{3}{uv}+\frac{15}{vw}=2, \\ \frac{15}{vw}+\frac{5}{wu}=2, \\ \frac{5}{wu}+\frac{3}{uv}=2 .\end{array}\right.$
Solution.

Domain of definition: $\left\{\begin{array}{l}u \neq 0, \\ v \neq 0, \\ w \neq 0 .\end{array}\right.$

Let $\frac{3}{u v}=x, \frac{15}{v w}=y, \frac{5}{w u}=z$.

With respect to $x, y$ and $z$, the system of equations becomes

$\left\{\begin{array}{l}x+y=2, \\ y+z=2, \\ z+x=2 .\end{array} \Rightarrow z=2-x\right.$.

Substituting this value of $z$ into the second equation, we get

$$
\left\{\begin{array} { l } 
{ x + y = 2 , } \\
{ y + 2 - x = 2 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ x + y = 2 , } \\
{ y - x = 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
2 y=2, \\
2 x=2 .
\end{array} \Rightarrow x=1, y=1\right.\right.\right.
$$

Then $z=2-1=1$.

Thus $\left\{\begin{array}{l}\frac{3}{u v}=1, \\ \frac{15}{v w}=1, \\ \frac{5}{w u}=1\end{array} \Leftrightarrow\left\{\begin{array}{l}u v=3, \\ v w=15 \\ w u=5\end{array}, \Rightarrow w=\frac{5}{u}\right.\right.$.

Substituting this value of $w$ into the second equation, we have

$$
\left\{\begin{array} { l } 
{ u v = 3 , } \\
{ v \cdot \frac { 5 } { u } = 1 5 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
u v=3, \\
v=3 u
\end{array} \Rightarrow u \cdot 3 u=3, u^{2}=1, \text { from which } u_{1}=-1, u_{2}=1 .\right.\right.
$$

Then $v_{1}=-3, v_{2}=3 ; w_{1}=-5, w_{2}=5$.

Answer: $(-1 ;-3 ;-5),(1 ; 3 ; 5)$.