The distance of the plane $\vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})=3$ from the origin is :

$\frac{3}{\sqrt{6}}$

$\vec{r} . \underbrace{(2 \hat{i}-\hat{j}+\hat{k})}=3$
    for a place $\vec{n}$

$\vec{r} . \hat{n}=d$   d → distance form origin

So here $|\vec{n}|=\sqrt{4+1}+1=\sqrt{6}$ (not a unit vector)

dividing equation by $|\vec{n}|$

$\vec{r} . \frac{(2 \hat{i}-\hat{j}+\hat{k})}{\sqrt{6}}=\frac{3}{\sqrt{6}}$

So  $\frac{3}{\sqrt{6}}$ → distance from origin