5.19. Given three non-zero vectors $\bar{a}, \bar{b}, \bar{c}$, each pair of which are non-collinear. Find their sum, if $(\bar{a}+\bar{b}) \| \bar{c}$ and $(\bar{b}+\bar{c}) \| \bar{a}$.
5.19. We have

$$
\begin{aligned}
& (\bar{a}+\bar{b}) \| \bar{c} \Rightarrow \bar{c}=m(\bar{a}+\bar{b})=m \bar{a}+m \bar{b} \\
& (\bar{b}+\bar{c}) \| \bar{a} \Rightarrow \bar{b}+\bar{c}=n \bar{a} \Rightarrow \bar{c}=n \bar{a}-\bar{b}
\end{aligned}
$$

Due to the uniqueness of the decomposition of $\bar{c}$ in terms of vectors $\bar{a}$ and $\bar{b}$, we conclude that $m=n$ and $m=-1$. Therefore, $m=n=-1$, i.e., $\bar{c} = -\bar{a}-\bar{b}$. Now we find $\bar{a}+\bar{b}+\bar{c}=\bar{a}+\bar{b}-\bar{a}-\bar{b}=\overline{0}$.

Answer: $\overline{0}$.