The distance of the plane \(\vec{r}\).\((\frac{2}{7}\hat{i}+\frac{3}{7}\hat{j}-\frac{6}{7}\hat{k}\))=2 from the origin is

2 unit 

$\vec{r}.(\frac{2}{7}\hat{i}+\frac{3}{7}\hat{j}-\frac{6}{7}\hat{k})=2$

Substituting $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$

$x\hat{i}+y\hat{j}+z\hat{k}.(\frac{2}{7}\hat{i}+\frac{3}{7}\hat{j}-\frac{6}{7}\hat{k})=2$

$=\frac{1}{7}[2x+3y+6z]=2⇒2x+3y+6z=2×7$

$=2x+3y+6z=14⇒2x+3y+6z-14=0$

Distance of plane $ax+by+cz+d=0$ from origin is given by -

$\frac{|d|}{\sqrt{a^2+b^2+c^2}}=\frac{|-14|}{\sqrt{2^2+3^2+6^2}}=\frac{14}{\sqrt{4+9+36}}⇒\frac{14}{\sqrt{49}}=\frac{14}{7}$ = 2 units

∴ option 2 is correct.