Let \(\overline{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overline{\mathrm{b}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}\) and \(\overline{\mathrm{c}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}\) be three vectors. A vector \(\overline{\mathrm{v}}\) in the plane of \(\overline{\mathrm{a}}\) and \(\overline{\mathrm{b}}\), whose projection on \(\overline{\mathrm{c}}\) is \(\frac{1}{\sqrt{3}}\), is given by

\(\bar{a}=\hat{i}+\hat{j}+\hat{k}, \bar{b}=\hat{i}-\hat{j}+\hat{k} \text { and } \bar{c}=\hat{i}-\hat{j}-\hat{k}\)
\(\bar{v}\) is in the plane of \(\overline{\mathrm{a}}\) and \(\overline{\mathrm{b}}\).
\(\begin{aligned}
& \overline{\mathrm{v}}=\mathrm{m} \overline{\mathrm{a}}+n \overline{\mathrm{~b}} \\
& \Rightarrow \overline{\mathrm{v}}=(m+n) \hat{\mathrm{i}}+(m-n) \hat{j}+(m+n) \hat{\mathrm{k}}...(i)
\end{aligned}\)
Projection of \(\bar{v}\) on \(\bar{c}=\frac{\bar{v} \cdot \bar{c}}{|\bar{c}|}=\frac{1}{\sqrt{3}}\)
\(\begin{aligned}
& \Rightarrow \frac{(m+n)(1)+(m-n)(-1)+(m+n)(-1)}{\sqrt{1+1+1}}=\frac{1}{\sqrt{3}} \\
& \Rightarrow-m+n=1 \\
& \Rightarrow \mathrm{n}=1+\mathrm{m} \\
& \mathrm{v}=(2 \mathrm{~m}+1) \hat{\mathrm{i}}-\hat{\mathrm{j}}+(2 \mathrm{~m}+1) \hat{\mathrm{k}} \quad \ldots[\text { From (i) }]
\end{aligned}\)
When \(\mathrm{m}=1\) then \(\overline{\mathrm{v}}=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}\)