6.203. $\left\{\begin{array}{l}x+y+z=0, \\ c x+a y+b z=0, \\ (x+b)^{2}+(y+c)^{2}+(z+a)^{2}=a^{2}+b^{2}+c^{2},\end{array} a \neq b \neq c\right.$.

6.203. $\left\{\begin{array}{l}x+y+z=0, \\ c x+a y+b z=0, \\ (x+b)^{2}+(y+c)^{2}+(z+a)^{2}=a^{2}+b^{2}+c^{2},\end{array} a \neq b \neq c\right.$.

The system of equations is:
\[
\left\{\begin{array}{l}
x + y + z = 0, \\
c x + a y + b z = 0, \\
(x + b)^{2} + (y + c)^{2} + (z + a)^{2} = a^{2} + b^{2} + c^{2},
\end{array} \quad a \neq b \neq c\right.
\]
## Solution.

Rewrite the equation as

$$
\begin{aligned}
& \left\{\begin{array}{l}
x+y+z=0, \\
c x+a y+b z=0, \\
x^{2}+2 b x+b^{2}+y^{2}+2 c y+c^{2}+z^{2}+2 a z+a^{2}=a^{2}+b^{2}+c^{2}
\end{array} \Leftrightarrow\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
x+y+z=0, \\
c x+a y+b z=0, \\
(x+y+z)^{2}-2 x y-2 x z-2 y z+2 b x+2 c y+2 a z=0
\end{array} \Leftrightarrow\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ x + y + z = 0 , } \\
{ c x + a y + b z = 0 , } \\
{ x y + y z + x z - b x - c y - a z = 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
z=-(x+y), \\
c x+a y+b z=0, \\
x y-b x-c y+(y+x-a) z=0
\end{array}\right.\right. \\
& \Rightarrow\left\{\begin{array}{l}
c x+c y-b(x+y)=0, \\
x y-b x-c y-(y+x-a)(x+y)=0
\end{array} \Rightarrow\right. \\
& \Rightarrow\left\{\begin{array}{l}
y=\frac{c-b}{b-a} x, \\
(x-c+a) y-y^{2}-(b+x-a) x=0 .
\end{array}\right.
\end{aligned}
$$

Substituting $y=\frac{c-b}{b-a} x$ from the second equation of the system into the third, we get

$$
\frac{(x-c+a)(c-b)}{b-a} x-\frac{(c-b)^{2}}{(b-a)^{2}} x^{2}-x^{2}-(b-a) x=0 \Rightarrow x(x-(a-b))=0
$$

from which $x_{1}=0, x_{2}=a-b ; y_{1}=0, z_{1}=0 ; y_{2}=\frac{c-b}{b-a} \cdot(a-b)=b-c$,

$$
z_{2}=-(a-b+b-c)=c-a
$$

Answer: $(0 ; 0 ; 0),(a-b ; b-c ; c-a)$.