Let $\mathit{a}\mathbf{,}\mathit{b}\mathbf{,}\mathit{c}$ be the three vectors such that  $\mathit{a}\mathbf{.}\mathbf{(}\mathit{b}\mathbf{+}\mathit{c}\mathbf{)}\mathbf{=}\mathit{b}\mathbf{.}\mathbf{(}\mathit{c}\mathbf{+}\mathit{a}\mathbf{)}\mathbf{=}\mathit{c}\mathbf{.}\mathbf{(}\mathit{a}\mathbf{+}\mathit{b}\mathbf{)}\mathbf{=}\mathbf{0}$ and $\left|\mathit{a}\right|=1$, $\left|\mathit{b}\right|=4$,  $\left|\mathit{c}\right|=8$, then $\left|\mathit{a}\mathbf{+}\mathit{b}\mathbf{+}\mathit{c}\right|$ = 

$\begin{array}{r}|\mathbf{a}+\mathbf{b}+\mathbf{c}{|}^2=(\mathbf{a}+\mathbf{b}+\mathbf{c})\cdot (\mathbf{a}+\mathbf{b}+\mathbf{c})\\ \mathbf{a}\cdot (\mathbf{b}+\mathbf{c})=\mathbf{b}\cdot (\mathbf{c}+\mathbf{a})=\mathbf{c}\cdot (\mathbf{a}+\mathbf{b})=0\\ 2(\mathbf{a}\cdot \mathbf{b}+\mathbf{b}\cdot \mathbf{c}+\mathbf{a}\cdot \mathbf{c})=0\end{array}$

$\begin{array}{r}\Rightarrow |\mathbf{a}+\mathbf{b}+\mathbf{c}{|}^2=|\mathbf{a}{|}^2+|\mathbf{b}{|}^2+|\mathbf{c}{|}^2\\ +2(\mathbf{a}\cdot \mathbf{b}+\mathbf{a}\cdot \mathbf{c}+\mathbf{b}\cdot \mathbf{c})\\ =1+16+64=81.\text{ Hence }|\mathbf{a}+\mathbf{b}+\mathbf{c}|=9.\end{array}$