Question

The equation of the planes parallel to the plane $x–2y+2z–3=0$ which are at unit distance from the point $(1,2,3)$ is $ax+by+cz+d=0$. If $(b–d)=K(c–a),$ then the positive value of $K$ is

Answer 1

$x–2y+2z+\lambda =0$
Now, given
$d=\frac{|1-4+6+\lambda |}{\sqrt{9}}=1$
$\Rightarrow |\lambda +3|=3$
$\Rightarrow \lambda +3=\pm 3\Rightarrow \lambda =0,-6$
So, planes are:
$P_1:x–2y+2z–6=0$
$P_2:x–2y+2z=0$
$\Rightarrow b–d=–2+6=4$
$\Rightarrow c–a=2-1=1$
$\Rightarrow \frac{b-d}{c-a}=K$
$\Rightarrow K=4$