3B. In the set of real numbers, solve the system of equations

$$
\left\{\begin{array}{l}
x+y+z=\frac{13}{3} \\
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{13}{3} \\
x y z=1
\end{array}\right.
$$
Solution. We have: $y+z=\frac{13}{3}-x, x(y+z)=\frac{13}{3}-y z, z y=\frac{1}{x}$. If we substitute from the first and third equation into the second, we get the equation

$$
x\left(\frac{13}{3}-x\right)=\frac{13}{3}-\frac{1}{x}
$$

which is equivalent to the equation $13 x^{2}-3 x^{3}-13 x+3=0$, i.e., to the equation

$$
(x-1)(3 x-10 x+3)=0
$$

The solutions to the last equation are $x_{1}=1, x_{2}=3, x_{3}=\frac{1}{3}$.

Since the equations in the system are symmetric, we conclude that the solutions to the system are the following six ordered triples, i.e., the permutations of the numbers: 1, 3, and $\frac{1}{3}$.