If \(\bar{a}, \bar{b}, \bar{c}\) are the three unit vectors such that \(|\bar{a}+\bar{b}+\bar{c}|=1\) and \(\bar{b}\) is perpendicular to \(\bar{c}\). If \(\bar{a}\) makes angles \(\alpha, \beta\) with \(\bar{b}\) and \(\bar{c}\) respectively, then \(\cos \alpha+\cos \beta\) has the value

\( |\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}|^2=|\overrightarrow{\mathrm{a}}|^2+|\overrightarrow{\mathrm{b}}|^2+|\overrightarrow{\mathrm{c}}|^2+2 \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}~+\) \(2 \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}+2 \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}} \)
\( \Rightarrow 1^2=1^2+1^2+1^2+2 \)
\( |\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \cos \alpha+2|\overrightarrow{\mathrm{b}}||\overrightarrow{\mathrm{c}}| \cos 90^{\circ}+2|\overrightarrow{\mathrm{c}}||\overrightarrow{\mathrm{a}}| \cos \beta \)
\( \Rightarrow 1+3+2 \cos \alpha+2 \times 0+2 \cos \beta \)
\( \Rightarrow \cos \alpha+\cos \beta=-1\)