Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Given equation of plane
$x-2y+2z+4=0$….(i)
Equation of plane parallel (ii) from point (1, 2, 3) is 1 unit
$\therefore \frac{\left|1\left(1\right)-2\left(2\right)+2\left(3\right)+k\right|}{\sqrt{{\left(1\right)}^2+{\left(-2\right)}^2+{\left(2\right)}^2}}=1$
$\Rightarrow \frac{\left|1-4+6+k\right|}{\sqrt{1+4+4}}=1$
$\Rightarrow \frac{\left|3+k\right|}{\sqrt{9}}=1$
$\Rightarrow \left|3+k\right|=3\Rightarrow 3+k=\pm 3$
$\therefore$$x-2y+2z+0=0$and$x-2y+2z-6=0$
$\Rightarrow x-2y+2z=0$and$x-2y+2z-6=0$