307. Find the sum of the vectors $\overline{a_{1}}=2 \bar{i}+3 \bar{j}-\bar{k}, \quad \bar{a}_{2}=3 \bar{k}-\bar{i}$, $\bar{a}_{3}=-\bar{i}+\bar{j}-2 \bar{k}$
Solution. $\quad \overline{a_{1}}+\overline{a_{2}}+\overline{a_{3}}=2 \bar{i}+3 \bar{j}-\bar{k}+3 \bar{k}-\bar{i}-\bar{i}+\bar{j}-2 \bar{k}=$ $=4 \bar{j}$. Let's denote the sum of vectors $\bar{a}_{1}+\bar{a}_{2}+\bar{a}_{3}$ by $\bar{a}$ and write the obtained result as follows:

$$
\bar{a}\{0 ; 4 ; 0\} \text { or } \bar{a}=\{0 ; 4 ; 0\}
$$

The vector $\bar{a}$ is collinear with the $O y$ axis.