If the equation of the plane which is at a di | vedclass

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(A) The direction ratios of the normal to the plane are $(1, 2, 2)$.Thus,the equation of the plane is $x+2y+2z=d$.Given that the distance from the ori

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$3$

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(A) The direction ratios of the normal to the plane are $(1, 2, 2)$.Thus,the equation of the plane is $x+2y+2z=d$.Given that the distance from the origin is $\frac{1}{3}$,we use the perpendicular distance formula:$\left|\frac{-d}{\sqrt{1^2+2^2+2^2}}\right| = \frac{1}{3} \Rightarrow \left|\frac{-d}{3}\right| = \frac{1}{3} \Rightarrow |d|=1$.Taking $d=1$,the equation of the plane is $x+2y+2z-1=0$.Comparing this with $x+py+qz+r=0$,we get $p=2, q=2, r=-1$.Therefore,$\sqrt{p^2+q^2+r^2} = \sqrt{2^2+2^2+(-1)^2} = \sqrt{4+4+1} = \sqrt{9} = 3$.

If the equation of the plane which is at a distance of $\frac{1}{3}$ units from the origin and perpendicular to a line whose directional ratios are $(1, 2, 2)$ is $x+py+qz+r=0$,then $\sqrt{p^2+q^2+r^2}=$

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