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

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 \((1,2,3)\)
\(\left|\frac{\mathrm{ax}_1+\mathrm{by}_1+\mathrm{cz} z_1+\mathrm{d}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}\right|=1 \)
\( \left|\frac{1(1)+(-2)(2)+2(3)+\lambda}{\sqrt{1+4+4}}\right|=1 \Rightarrow\left|\frac{1-4+6+\lambda}{3}\right|=\) \(1 \Rightarrow 3+\lambda= \pm 3 \)
\( \therefore 3+\lambda=3 \text { or } 3+\lambda=-3 \)
\( \lambda=0 \text { or } \lambda=-6\)
The equation of planes are \(x-2 y+2 z=0\) and \(x-2 y+2 z-6=0\)