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

$\vec a=\hat i+\hat j+\hat k$

$\vec b=\hat i-\hat j+\hat k$

$\vec c=\hat i-\hat j-\hat k$

$\vec v \in \text{span}\{\vec a,\vec b\}$

$\text{proj}_{\vec c}\,\vec v = \dfrac{\vec v\cdot\vec c}{|\vec c|} = \dfrac{1}{\sqrt3}$

$\text{To find: } \vec v$

$\vec v\in\text{span}\{\vec a,\vec b\} \iff \vec v\cdot(\vec a\times\vec b)=0$ $\vec a\times\vec b = \big(1\cdot1-1\cdot(-1),\ -(1\cdot1-1\cdot1),\ 1\cdot(-1)-1\cdot1\big)$ $= (2,0,-2) \parallel (1,0,-1)$ $\vec v\cdot(1,0,-1)=0 \Rightarrow v_x = v_z$ $|\vec c| = \sqrt{1+1+1} = \sqrt3$ $\text{proj} = \dfrac{\vec v\cdot\vec c}{|\vec c|} = \dfrac{1}{\sqrt3} \Rightarrow \vec v\cdot\vec c = 1$ $\vec v\cdot\vec c = v_x - v_y - v_z = -v_y\ \ (v_x=v_z)$ $\Rightarrow -v_y = 1 \Rightarrow v_y = -1,\quad v_x = v_z$ $\vec v = (3,-1,3) = \vec a + 2\vec b \ \checkmark$ $\boxed{\text{option (3)}}$