306. Find the coordinates $X, Y, Z$ of the sum of vectors $\bar{a}\{2 ;-3$; $4\}, \bar{b}\{-5 ; 1 ; 6\}, \bar{c}=\{3 ; 0 ;-2\}$.
Solution. $X=2-5+3=0 ; \quad Y=-3+1+0=-2$; $Z=4+6-2=8$

Therefore, the sum of the vectors $\bar{a}+\bar{b}+\bar{c}=\{0 ;-2 ; \quad 8\}$. The required vector is parallel to the $y O z$ plane, since its component along the $O x$ axis is zero.