Let \(\mathbf{a}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{c}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}\). A vector coplanar to \(\mathbf{a}\) and \(\mathbf{b}\) has a projection along c of magnitude \(\frac{1}{\sqrt{3}}\), then the vector is

Let vector \(\mathbf{r}\) be coplanar to \(\mathbf{a}\) and \(\mathbf{b}\).
\[
\begin{array}{ll}
\therefore & \mathbf{r}=\mathbf{a}+t \mathbf{b} \\
\Rightarrow & \mathbf{r}=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})+t(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})=\hat{\mathbf{i}}(1+t)+\hat{\mathbf{j}}(2-t)+\hat{\mathbf{k}}(1+t)
\end{array}
\]
The projection of \(\mathbf{r}\) on \(\mathbf{c}=\frac{1}{\sqrt{3}} \Rightarrow \frac{\mathbf{r} \cdot \mathbf{c}}{|\mathbf{c}|}=\frac{1}{\sqrt{3}}\) \(\Rightarrow \frac{|1 \cdot(1+t)+1 \cdot(2-t)-1 \cdot(1+t)|}{\sqrt{3}}=\frac{1}{\sqrt{3}}\) \(\Rightarrow|2-t|=\pm 1 \Rightarrow t=1\) or 3
When, \(t=1\) we have \(\mathbf{r}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}\)
When, \(t=3\) we have \(\mathbf{r}=4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}}\)