The vector \(\overline{\mathrm{a}}=\alpha \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\beta \hat{\mathrm{k}}\) lies in the plane of the vectors \(\bar{b}=\hat{i}+\hat{j}\) and \(\bar{c}=\hat{j}+\hat{k}\) and bisects the angle between \(\bar{b}\) and \(\bar{c}\). Then which one of the following gives possible values of \(\alpha\) and \(\beta\) ?

Since \(\overline{\mathrm{a}}\) bisects the angle between \(\overline{\mathrm{b}}\) and \(\overline{\mathrm{c}}\).
\(\therefore \quad\) The equation of bisector of \(\overline{\mathrm{b}}\) and \(\overline{\mathrm{c}}\) is
\(\begin{aligned}
& \overline{\mathrm{a}}=\lambda(\hat{\mathrm{b}}+\hat{\mathrm{c}}) \\
& \Rightarrow \alpha \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\beta \hat{\mathrm{k}}=\lambda\left(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}+\frac{\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{2}}\right)
\end{aligned}\)
\(\Rightarrow \alpha \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\beta \hat{\mathrm{k}}=\frac{\lambda}{\sqrt{2}}(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})\)
On comparing, we get \(\alpha=\frac{\lambda}{\sqrt{2}}, 2=\sqrt{2} \lambda\) and \(\beta=\frac{\lambda}{\sqrt{2}}\) \(\Rightarrow \alpha=1, \beta=1\)