If \(\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}\) are three vectors such that \(|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|=1\), \(\overline{\mathrm{c}}=\lambda(\overline{\mathrm{a}} \times \overline{\mathrm{b}})\) and \(|\overline{\mathrm{a}}|=\frac{1}{\sqrt{3}},|\overline{\mathrm{b}}|=\frac{1}{\sqrt{2}},|\overline{\mathrm{c}}|=\frac{1}{\sqrt{6}}\), then the angle between \(\bar{a}\) and \(\bar{b}\) is

Let \(\theta\) be the angle between \(\bar{a}\) and \(\bar{b}\).
\(
\begin{aligned}
& \text { Since } \overline{\mathrm{c}}=\lambda(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \\
& \Rightarrow \overline{\mathrm{c}} \perp \overline{\mathrm{a}}, \overline{\mathrm{c}} \perp \overline{\mathrm{b}} \\
& \Rightarrow \overline{\mathrm{c}} \cdot \overline{\mathrm{a}}=\overline{\mathrm{c}} \cdot \overline{\mathrm{b}}=0
\end{aligned}
\)
Now,
\( |\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|=1 \)
\( \Rightarrow|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|^2=1 \)
\( \Rightarrow|\overline{\mathrm{a}}|^2+|\overline{\mathrm{b}}|^2+|\overline{\mathrm{c}}|^2+2(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}+\overline{\mathrm{b}}\) \(\cdot \overline{\mathrm{c}}+\overline{\mathrm{c}} \cdot \overline{\mathrm{a}})=1 \)
\( \Rightarrow \frac{1}{3}+\frac{1}{2}+\frac{1}{6}+2\{|\overline{\mathrm{a}}||\overline{\mathrm{b}}| \cos \theta\}=1 \)
\( \Rightarrow \cos \theta=0 \)
\( \Rightarrow \theta=\frac{\pi}{2}\)