Let $\overrightarrow d=x\hat i+y\hat j+z\hat k$ be vector in the plane of $\overrightarrow a\:and\:\overrightarrow b$

$\Rightarrow\:[\overrightarrow a\:\overrightarrow b\:\overrightarrow d]=0$

$\Rightarrow\:(\overrightarrow a\times\overrightarrow b).\overrightarrow d=0$

$\Rightarrow\:3x-3z=0,\:\:\Rightarrow\:x-z=0.........(i)$

Given projection of $\overrightarrow d$ in $\overrightarrow c =\large\frac{1}{\sqrt 3}$

$\Rightarrow\:\large\frac{\overrightarrow d.\overrightarrow c}{|\overrightarrow c|}=\frac{1}{\sqrt3}$

$\Rightarrow\:\large\frac{x+y-z}{\sqrt 3}=\frac{1}{\sqrt 3}$

$\Rightarrow\:x+y-z=1............(ii)$

Solving (i) and (ii) we get

$x=z$ and $y=1$

which satisfies the vector $-2\hat i+\hat j-2\hat k$