If \(\bar{a}=\hat{i}+2 \hat{j}+\hat{k}, \bar{b}=\hat{i}-\hat{j}+\hat{k}, \bar{c}=\hat{i}+\hat{j}-\hat{k}\), then a vector in the plane of \(\bar{a}\) and \(\bar{b}\), whose projection on \(\overline{\mathrm{c}}\) is \(\frac{1}{\sqrt{3}}\), is

Let \(\overline{\mathrm{r}}\) be the vector coplanar to \(\overline{\mathrm{a}}\) and \(\overline{\mathrm{b}}\). Then,
\(\begin{aligned}
\overline{\mathrm{r}} & =\overline{\mathrm{a}}+\mathrm{m} \overline{\mathrm{b}} \\
& =(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})+\mathrm{m}(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}) \\
& =\hat{\mathrm{i}}(1+\mathrm{m})+\hat{\mathrm{j}}(2-\mathrm{m})+\hat{\mathrm{k}}(1+\mathrm{m})
\end{aligned}\)
Since the projection of \(\overline{\mathrm{r}}\) along \(\overline{\mathrm{c}}\) is \(\frac{1}{\sqrt{3}}\),
\(\begin{aligned}
& \frac{\overline{\mathrm{r}} \cdot \overline{\mathrm{c}}}{\overline{\mathrm{c}}}= \pm \frac{1}{\sqrt{3}} \\
& \Rightarrow \frac{(1+\mathrm{m})+(2-\mathrm{m})-(1+\mathrm{m})}{\sqrt{3}}= \pm \frac{1}{\sqrt{3}} \\
& \Rightarrow(1+\mathrm{m})+(2-\mathrm{m})-(1+\mathrm{m})= \pm 1 \\
\therefore \quad & \mathrm{m}=3 \text { or } \mathrm{m}=1
\end{aligned}\)
Substituting \(\mathrm{m}=3\) in equation (i), we get
\(\begin{aligned}
& \overline{\mathrm{r}}=\hat{\mathrm{i}}(1+3)+\hat{\mathrm{j}}(2-3)+\hat{\mathrm{k}}(1+3) \\
& \Rightarrow \overline{\mathrm{r}}=4 \hat{\mathrm{i}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}}
\end{aligned}\)