\(\bar{a}, \bar{b}, \bar{c}\) are vectors such that \(|\bar{a}|=5,|\bar{b}|=4,|\bar{c}|=3\) and each is perpendicular to the sum of the other two, then \(|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|^2=\)

We have \(\overline{\mathrm{a}} \cdot(\overline{\mathrm{b}}+\overline{\mathrm{c}})=0 \cdot \overline{\mathrm{b}} \cdot(\overline{\mathrm{c}}+\overline{\mathrm{a}})=0\) and \(\overline{\mathrm{c}} \cdot(\overline{\mathrm{a}}+\overline{\mathrm{b}})=0\)
\(
\begin{aligned}
\therefore ~& \overline{\mathrm{a}} \cdot \overline{\mathrm{b}}+\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}=0 \\
& \overline{\mathrm{b}} \cdot \overline{\mathrm{c}}+\overline{\mathrm{b}} \cdot \overline{\mathrm{a}}=0 \\
& \overline{\mathrm{c}} \cdot \overline{\mathrm{a}}+\overline{\mathrm{c}} \cdot \overline{\mathrm{b}}=0
\end{aligned}
\)
From (1), (2) and (3), we get
\(2(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}+\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}+\overline{\mathrm{c}} \cdot \overline{\mathrm{a}})=0 \)
\( \text {Now }|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|^2=|\overline{\mathrm{a}}|^2+|\overline{\mathrm{b}}|^2+|\overline{\mathrm{c}}|^2+2(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}+\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}\) \(+~\overline{\mathrm{c}} \cdot \overline{\mathrm{a}}) \)
\( \therefore|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|^2=(5)^2+(4)^2+(3)^2+2(0)=50\)