3. (7p) Solve in $\boldsymbol{R} \times \boldsymbol{R} \times \boldsymbol{R}$ the system:

$$
\left\{\begin{array}{l}
2^{x}+3^{x}=3 y+2 \\
2^{y}+3^{y}=3 z+2 \\
2^{z}+3^{z}=3 x+2
\end{array}\right.
$$

Petru Vlad
3.Systolic. Assume that ( $x, y, z$ ) is a solution of the system and $x>y$ $x>y \Rightarrow 3 y+2=2^{x}+3^{x}>2^{y}+3^{y}=3 z+2 \Rightarrow y>z$.....

(1p)

$3 z+2=2^{y}+3^{y}>2^{z}+3^{z}=3 x+2 \Rightarrow z>x$,

contradiction.....

Similarly for $x<y$, so $x=y$, and then

$x=y=z$.

We have that

$2^{x}+3^{x}=3 x+2$.

But $f: R \rightarrow R, f(x)=2^{x}+3^{x}$ is a convex function and the geometric representation of the graph of the function $g: R \rightarrow R, g(x)=3 x+2$ is a straight line. Using graphical reading, we deduce that the intersection is possible at most in two distinct points
$x=0, x=1$ satisfy the equation, so $S=\{(0,0,0),(1,1,1)\}$.