3. Given are positive numbers $a, b$ and $c$. Determine all triples of positive numbers $(x, y, z)$ such that

$$
x+y+z=a+b+c \quad \text { and } \quad 4 x y z-a^{2} x-b^{2} y-c^{2} z=a b c .
$$
3. Let's write the second equation in the form $\frac{a^{2}}{y z}+\frac{b^{2}}{z x}+\frac{c^{2}}{x y}+\frac{a b c}{x y z}=4, \text{i.e.}$.

$$
x_{1}^{2}+y_{1}^{2}+z_{1}^{2}+x_{1} y_{1} z_{1}=4, \quad \text{where} \quad x_{1}=\frac{a}{\sqrt{y z}}, \quad y_{1}=\frac{b}{\sqrt{z x}} \text{ and } z_{1}=\frac{c}{\sqrt{x y}}
$$

It is clear that $0 < x, y, z$ and if $x+y+z=a+b+c$, then $f(x, y, z)=4 x y z-a^{2} x-b^{2} y-c^{2} z \leqslant a b c$. If $(x, y, z)$ is a point of local extremum of the function $f$ in the domain $x, y, z > 0$ under the condition $g(x, y, z)=x+y+z-a+b+c=0$, then by the theorem on Lagrange multipliers, there exists a real number $\lambda$ such that $(x, y, z)$ is a stationary point of the function $F(x, y, z)=f-\lambda g$, i.e., $\frac{\partial F}{\partial x}=\frac{\partial F}{\partial y}=\frac{\partial F}{\partial z}=0$.

We have $\frac{\partial F}{\partial x}=4 y z-a^{2}-\lambda$ and so on. It follows that $4 y z-a^{2}=4 z x-b^{2}=4 x y-c^{2}=\lambda$ and $x+y+z=a+b+c$. The only solution to this system is $(x, y, z)=\left(\frac{b+c}{2}, \frac{c+a}{2}, \frac{a+b}{2}\right)$. Finally, we need to examine the boundary of the domain $D$, i.e., when one of the numbers $x, y, z$ is zero. Then, of course, $F<0$. It follows that in the given domain $\max F(x, y, z)=F\left(\frac{b+c}{2}, \frac{c+a}{2}, \frac{a+b}{2}\right)=a b c$.