If $\overrightarrow{\mathrm{\alpha }}+\overrightarrow{\mathrm{\beta }}+\overrightarrow{\mathrm{\gamma }}=\mathrm{k}\overrightarrow{\mathrm{\delta }}\text{ and }\overrightarrow{\mathrm{\beta }}+\overrightarrow{\mathrm{\gamma }}+\overrightarrow{\mathrm{\delta }}=m\overrightarrow{\mathrm{\alpha }},\overrightarrow{\mathrm{\alpha }}\text{ and }\overrightarrow{\mathrm{\delta }}$  are non-collinear, then $\overrightarrow{\mathrm{\alpha }}+\overrightarrow{\mathrm{\beta }}+\overrightarrow{\mathrm{\gamma }}+\overrightarrow{\mathrm{\delta }}$ equal

$\text{ Given }\overrightarrow{\alpha }+\overrightarrow{\beta }+\overrightarrow{\gamma }=k\overrightarrow{\delta }----\left(i\right)$

$\overrightarrow{\mathrm{\beta }}+\overrightarrow{\mathrm{\gamma }}+\overrightarrow{\mathrm{\delta }}=\mathrm{m}\overrightarrow{\mathrm{\alpha }}----\left(\mathrm{ii}\right)$

From(i), $\overrightarrow{\mathrm{\alpha }}+\overrightarrow{\mathrm{\beta }}+\overrightarrow{\mathrm{\gamma }}+\overrightarrow{\mathrm{\delta }}=(\mathrm{k}+1)\overrightarrow{\mathrm{\delta }}---\left(\mathrm{iii}\right)$

From (ii), $\overrightarrow{\mathrm{\alpha }}+\overrightarrow{\mathrm{\beta }}+\overrightarrow{\mathrm{\gamma }}+\overrightarrow{\mathrm{\delta }}=(\mathrm{m}+1)\overrightarrow{\mathrm{\alpha }}----\left(\mathrm{iv}\right)$

From (iii) and (iv), we get 

$(\mathrm{k}+1)\overrightarrow{\mathrm{\delta }}=(\mathrm{m}+1)\overrightarrow{\mathrm{\alpha }}-----\left(\mathrm{v}\right)$

Since $\overrightarrow{\mathrm{\alpha }}$ is not parallel to $\overrightarrow{\mathrm{\delta }}$, 

From (v),k+1=0and m+ 1=0 

$\text{ From (iii), }\overrightarrow{\mathrm{\alpha }}+\overrightarrow{\mathrm{\beta }}+\overrightarrow{\mathrm{\gamma }}+\overrightarrow{\mathrm{\delta }}=\overrightarrow{0}$