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