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

Any plane parallel to the plane $x+2y+2z+8=0$ is given by
$x+2y+2z+\lambda =0$ ... (i),

where $\lambda$ is scalar.
Its distance from the point $(1,1,2)$
$=\pm \frac{1+2\left(1\right)+2\left(2\right)+\lambda }{\sqrt{1^2+2^2+2^2}}$
$=\pm \frac{\lambda +7}{3}$
According to the question, we have
$\pm \frac{\lambda +7}{3}=2$
$\therefore \frac{\lambda +7}{3}=\pm 2$
$\Rightarrow \lambda +7=\pm 6$
$\Rightarrow \lambda +7=6$ or $\lambda +7=-6$
$\Rightarrow \lambda =-1$ or $\lambda =-13$
From equation (i),
$x+2y+2z-1=0$
and $x+2y+2z-13=0$