Given,

Direction ratio of perpendicular line to the plane is $\left(1,2,2\right)$,

So the equation of the plane will be $x+2y+2z=d$

Now given distance from origin is $\frac{1}{3}$,

So by perpendicular distance formula we get,

$\left|\frac{d}{\sqrt{1^2+2^2+2^2}}\right|=\frac{1}{3}\Rightarrow d=1$

So, equation of plane will be $x+2y+2z-1=0$

Now comparing equation of plane with $x+py+qz+r=0$

We get, $p=2, q=2 \& r=-1$

So, the value of $\sqrt{p^2+q^2+r^2}=\sqrt{2^2+2^2+1^2}=\sqrt{9}=3$