Consider $\overrightarrow{\mathit{a}}\mathit{+}\overrightarrow{\mathit{b}}\mathit{+}\overrightarrow{\mathit{c}}\mathit{=}\overrightarrow{\mathit{k}}$.

Taking cross product of the above equation with$\overrightarrow{\mathit{a}}$,

$\overrightarrow{a}\mathit{\times }\left(\overrightarrow{a}\mathit{+}\overrightarrow{b}\mathit{+}\overrightarrow{c}\right)\mathit{=}\overrightarrow{a}\mathit{\times }\overrightarrow{k}$

$\Rightarrow \overrightarrow{a}\mathit{\times }\overrightarrow{a}\mathit{+}\overrightarrow{a}\mathit{\times }\overrightarrow{b}\mathit{+}\overrightarrow{a}\mathit{\times }\overrightarrow{c}\mathit{=}\overrightarrow{a}\mathit{\times }\overrightarrow{k}$

$\Rightarrow \overrightarrow{a}\mathit{\times }\overrightarrow{b}\mathit{-}\overrightarrow{c}\mathit{\times }\overrightarrow{a}\mathit{=}\overrightarrow{a}\mathit{\times }\overrightarrow{k}$

$\Rightarrow \overrightarrow{a}\mathit{\times }\overrightarrow{b}\mathit{-}\overrightarrow{a}\mathit{\times }\overrightarrow{b}\mathit{=}\overrightarrow{a}\mathit{\times }\overrightarrow{k}$

$\mathit{\Rightarrow }\mathit{0}\mathit{=}\overrightarrow{a}\mathit{\times }\overrightarrow{k}$

$\mathit{\Rightarrow }\mathit{0}\mathit{=}\overrightarrow{k}$

Therefore, _{$\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0$.}