Question

Find the equations of the planes parallel to the plane $x-2y+2z-4=0$, which are at a unit distance from the point $(1,2,3)$.

Answer 1

Any plane || to the plane $x-2y-2z-4=0$ can be written as,

$x-2y+2z-\lambda =0$ ..... (i), where $\lambda$ is a scalar quantity

Now, distance of the plane from the point $(1,2,3)$

$=\pm \frac{1\left(1\right)-2\left(2\right)+2\left(3\right)+\lambda }{\sqrt{1^2+(-2{)}^2+2^2}}$

$=\pm \frac{1-4+6+\lambda }{\sqrt{1+4+4}}=\pm \frac{3+\lambda }{\sqrt{9}}=\pm \frac{3+\lambda }{3}$

But $\pm \frac{3+\lambda }{3}=1$

$\Rightarrow 3+\lambda =\pm 3$

$\Rightarrow \lambda =0,-6$

Substituting the values of $\lambda$ in (i), we have

$x-2y+2z-(-6)=0$

$x-2y+2z+6=0$