Equation of planes parallel to the plane $x-2 y+2 z+4=0$ which are at a distance of one unit from the point…
Equation of planes parallel to the plane $x-2 y+2 z+4=0$ which are at a distance of one unit from the point $(1,2,3)$ are
$x+2 y+2 z=6, x+2 y+2 z=0$
$x-2 y+2 z=0, x-2 y+2 z-6=0$
$x-2 y-6=0, x-2 y+z=6$
$x+2 y+2 z=-6, x+2 y+2 z=5$
Solution
The equation of planes parallel to the plane $x-2 y+2 z+4=0$ is
$\mathrm{x}-2 \mathrm{y}+2 \mathrm{z}+\lambda=0$
The required planes are at a distance of one unit from $(1,2,3)$
$\begin{aligned}
& \left|\frac{\mathrm{ax}_1+\mathrm{by}_1+\mathrm{cz} z_1+\mathrm{d}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}\right|=1 \\
& \left|\frac{1(1)+(-2)(2)+2(3)+\lambda}{\sqrt{1+4+4}}\right|=1 \Rightarrow\left|\frac{1-4+6+\lambda}{3}\right|=1 \Rightarrow 3+\lambda= \pm 3 \\
& \therefore 3+\lambda=3 \quad \text { or } \quad 3+\lambda=-3 \\
& \lambda=0 \quad \text { or } \quad \lambda=-6
\end{aligned}$
The equation of planes are $x-2 y+2 z=0$ and $x-2 y+2 z-6=0$