Three Dimensional Geometry Part 2

Equation of any plane parallel to the plane x−2y+2z−3=0 is
x−2y+2z+λ=0....(i)
Given, distance from (1,2,3) is 1.
∴‌‌|‌

1−2×2+2×3+λ

√(1)2+(−2)2+(2)2

|=1
⇒‌‌|λ+3|=3
⇒‌‌λ+3=±3
⇒‌‌λ=0,−6
Consider, λ=−6
∴ Equation of required plane is
x−2y+2z−6=0
On comparing this equation to
ax+by+cz+d=0, we get
a=1,b=−2,c=2 and d=−6
∵‌‌(b−d)=k(c−a)
∴‌‌4=k×(1)
⇒ k=4