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

Given that,
\[
\begin{array}{l}
\vec{a}=\hat{i}+2 \hat{j}+\hat{k} \rightarrow(1) \\
\vec{b}=\hat{i}-\hat{j}+\hat{k} \rightarrow(2) \\
\vec{C}=\hat{i}+\hat{j}-\hat{k} \rightarrow(3)
\end{array}
\]
A vector in the plane of \( \vec{a} \) and \( \vec{b} \) can be written as,
\[
\begin{array}{l}
\vec{r}=m \vec{a}+n \vec{b} \\
=m(\hat{i}+2 \hat{j}+\hat{k})+n(\hat{i}-\hat{j}+\hat{k}) \\
\vec{r}=(m+n) \hat{i}+(2 m-n) \hat{j}+(m+n) \hat{k}
\end{array}
\]
So, projection on \( \vec{C} \) by \( \vec{r} \) is given by
\[
\begin{array}{l}
\frac{\vec{r} \cdot \vec{C}}{|\vec{C}|}=\frac{(m+n)+(2 m-n)-(m+n)}{\sqrt{1+1+1}} \\
\Rightarrow \frac{2 m-n}{\sqrt{3}}=\frac{1}{\sqrt{3}}
\end{array}
\]
As use can see from Eq. (4), the \( ^{\wedge} i \) component and the \( \hat{k} \) component must be equal and \( \hat{j} \) can be either \( -1 \) or \( 1 . \)
Therefore, option (4) is correct, that is \( 4 \hat{i}-\hat{j}+4 \hat{k} \)