6.224. $\left\{\begin{array}{l}\sqrt{x+y}+\sqrt{y+z}=3, \\ \sqrt{y+z}+\sqrt{z+x}=5, \\ \sqrt{z+x}+\sqrt{x+y}=4 .\end{array}\right.$
## Solution.

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

By adding all three equations of the system, we get

$$
2 \sqrt{x+y}+2 \sqrt{y+z}+2 \sqrt{z+x}=12 \Leftrightarrow \sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}=6
$$

By subtracting the first, second, and third equations of the system from the obtained equation, we have

$$
\left\{\begin{array} { l } 
{ \sqrt { z + x } = 3 } \\
{ \sqrt { x + y } = 1 , } \\
{ \sqrt { y + z } = 2 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ z + x = 9 } \\
{ x + y = 1 , } \\
{ y + z = 4 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=3 \\
y=-2 \\
z=6
\end{array}\right.\right.\right.
$$

Answer: $(3 ;-2 ; 6)$.