Let $\vec a=2\hat i-\hat j+\hat k,\vec b=\hat i+2\hat j-\hat k$ and $\vec c=\hat i+\hat j-2\hat k$. A vector in the plane of $\vec b$ and $\vec c$ whose projection on $\vec a$ has the magnitude $\sqrt{\frac{2}{3}}$ is

$-2\hat i-\hat j+5\hat k$

Let $\vec r$ be the required vector. Then, $\vec r =x\vec b+y\vec c$.

We have,

Projection of $\vec r$ on $\vec a =\sqrt{\frac{2}{3}}$

$⇒\vec r.\hat a=\sqrt{\frac{2}{3}}$

$⇒(x\vec b+y\vec c).\hat a=\sqrt{\frac{2}{3}}$

$⇒x(\vec b.\hat a)+y(\vec c.\hat a)=\sqrt{\frac{2}{3}}$

$⇒-\frac{x}{\sqrt{6}}-\frac{y}{\sqrt{6}}=\sqrt{\frac{2}{3}}⇒x+y=-2⇒y= -x-2$

Now,

$\vec r=x\vec b+y\vec c$

$⇒\vec r= (x + y)\hat i + (2x + y)\hat j -(x+2y)\hat k$

$⇒\vec r=-2\hat i+(x-2)\hat j+(x+4)\hat k$

Clearly, option (3) satisfies this for x = 1.