3. Solve the system of equations

$$
\begin{aligned}
a+b & =8 \\
a^{2}+b^{2}+c^{2} & =32
\end{aligned}
$$

in the set of real numbers.
3. From the first equation, it follows that $b=8-a$, so by substituting this into the second, we get

$$
32=a^{2}+(8-a)^{2}+c^{2}=a^{2}+64-16 a+a^{2}+c^{2}=2 a^{2}-16 a+64+c^{2}
$$

i.e., $2 a^{2}-16 a+32+c^{2}=0$. However, notice that the left side can be transformed as $2 a^{2}-16 a+32+c^{2}=2\left(a^{2}-8 a+16\right)+c^{2}=2(a-4)^{2}+c^{2}$, so for this to be equal to zero, it must be that $a-4=0$ and $c=0$. Therefore, the only solution to the given system is $a=4, b=8-a=4$ and $c=0$.