Question:
The equation of the planes parallel to the plane $x-2 y+2 z-3=0$ which are at unit distance from the point $(1,2,3)$ is $a x+b y+c z+d=0$. If $(b-d)=K(c-a)$, then the positive value of $K$ is
Solution:
Let plane is $x-2 y+2 z+\lambda=0$
distance from $(1,2,3)=1$
$\Rightarrow \frac{|\lambda+3|}{5}=1 \Rightarrow \lambda=0,-6$
$\Rightarrow a=1, b=-2, c=2, d=-6$ or 0
$\mathrm{b}-\mathrm{d}=4$ or $-2, \mathrm{c}-\mathrm{a}=1$
$\Rightarrow \mathrm{k}=4$ or $-2$
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