Three vectors $\overline{a},\overline{b},\overline{c}$   are such that $\overline{a}\times \overline{b}=3(\overline{a}\times \overline{c})$ . Also  $\left|\overline{a}\right|=\left|\overline{b}\right|=1$ , $\left|\overline{c}\right|=\frac{1}{3}$  . If the angle between  $\overline{b}$  and  $\overline{c}$  is  ${60}^0$ , then

$\overrightarrow{a}\times \overrightarrow{b}=3(\overrightarrow{a}\times \overrightarrow{c}) \overrightarrow{a}\times (\overrightarrow{b}-3\overrightarrow{c})=\overline{O} \Rightarrow \overrightarrow{a}\left|\right|(\overrightarrow{b}-3\overrightarrow{c}) \Rightarrow \overrightarrow{b}-3\overrightarrow{c}=\lambda \overrightarrow{a}$

$$\begin{gathered} {|\overline{b}-3\overline{c}|}^2={\lambda }^2{\left|\overline{a}\right|}^2 \\ \Rightarrow {\left|\overline{b}\right|}^2+9{\left|\overline{c}\right|}^2-6(\overline{b}.\overline{c})={\lambda }^2{\left|\overline{a}\right|}^2 \\ \Rightarrow 1^2+9{\left(\frac{1}{3}\right)}^2-6\left|\overline{b}\right|\left|\overline{c}\right|\cos {60}^0={\lambda }^2{\left(1\right)}^2 \end{gathered}$$

$$\begin{gathered} 1+9\times \frac{1}{9}-\mathrm{6.1.}\frac{1}{3}\times \frac{1}{2}={\lambda }^2 \\ \Rightarrow 1+1-1={\lambda }^2 \\ \Rightarrow {\lambda }^2=1 \Rightarrow \lambda =\pm 1 \\ \Rightarrow \overline{b}-3\overline{c}=1.\overline{a}\left(or\right)\overline{b}-3\overline{c}=-\overline{a} \end{gathered}$$

$\overline{a}=\overline{b}-3\overline{c}\Rightarrow \overline{b}=3\overline{c}+\overline{a}$
$\overline{-a}=\overline{b}-3\overline{c}\Rightarrow \overline{b}=3\overline{c}-\overline{a}$