Let \(\bar{a}=\hat{i}+\hat{j}+\widehat{k}, \bar{b}=\hat{i}-\hat{j}+\widehat{k}\) and \(\bar{c}=\hat{i}-\hat{j}-\widehat{k}\) 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

Any vector in the plane of \(\vec{a}\) and \(\vec{b}\) can be written as
\(
\vec{v}=\vec{a}+\lambda \vec{b}=(\hat{i}+\hat{j}+\widehat{k})+\lambda(\hat{i}-\hat{j}+\widehat{k})
\)

\(\frac{\mathrm{A}}{\mathrm{Q}}\) Its projection on \(\overrightarrow{\mathrm{c}}\) is \(\frac{1}{\sqrt{3}}\)
\(
\begin{aligned}
& \Rightarrow \frac{(\vec{a}+\lambda \vec{b}) \cdot \vec{c}}{|\vec{c}|}=\frac{1}{\sqrt{3}} \\
& \Rightarrow \frac{(1+\lambda)-(1-\lambda)-(1+\lambda)}{\sqrt{1^2+(-1)^2+(-1)^2}}=\frac{1}{\sqrt{3}} \\
& \Rightarrow \lambda=2
\end{aligned}
\)
Putting \(\lambda=2\) in eq. (1) we get
\(
\vec{v}=3 \hat{i}-2 \hat{j}+3 \widehat{k}
\)