Question

Find the equations of the planes parallel to the plane $x+2y+2z+8=0$ which are at the distance of 2 units from the point (1, 1, 2).

Answer 1

We know that

Equation of planes parallel to given plane $x+2y+2z+8=0$ is

$x+2y+2x+\lambda =\mathrm{0....}\left(1\right)$

where ${\lambda }^{-}$ is a variable

Now this plane is at $2$ unit

distance from point $P(1,1,2)$

$D=\left|\frac{a{x}_1+b{y}_1+c{z}_1+d}{\sqrt{a_1^2+b_1^2+c_1^2}}\right|$

$2=\left|\frac{1\left(1\right)+2\left(1\right)+2\left(2\right)+\lambda }{\sqrt{(1{)}^2+(2{)}^2+(2{)}^2}}\right|$

$2=\left|\frac{7+\lambda }{\sqrt{9}}\right|$

Hence

$\lambda =(-7\pm 6)=-13,-1$

So equation of required plane is

$x+2y+2z-13=0---\left(A\right)$

$x+2y+2z-1=0----\left(B\right)$