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 The equation of planes parallel to the plane x+2y+2z+8=0 , which are at a distance of 2 units from the point (1,1,2) are

Answer:

Explanation:

 Equation  of plane parallel  to the plane

x+2y+2z+8=0 is x+2y+2z+λ=0

 Given distance from the  point (1,1,2)  to the plane.

 x+2y+2z +$\lambda$=0 is 2 units

$\therefore$             $2=|\frac{1+2+4+\lambda}{\sqrt{1+4+4}}|$

$\Rightarrow$     $ \pm2=\frac{\lambda+7}{3}$

$\Rightarrow \lambda+7=\pm6$

 $\therefore$      $\lambda$  = -13,-1

 Hence , equation of plane is

x+2y+2z-13=0 or x+2y+2z-1=0