The equation of the plane passes through the point (2,3,−1) is a(x−2)+b(y−3)+c(z+1)=0 .......(i) where a,b,c are the direction ratio of the normal to the plane. Also, given the plane is perpendicular to the line whose direction ratio is (3,−4,7). So, that line and the normal of the plane are parallel. ⇒
a
3
=
b
−4
=
c
7
=k ⇒ a=3k,b=−4k,c=7k From Eq. (i) 3k(x−2)−4k(y−3)+7k(z+1)=0 ⇒ 3x−4y+7z+13=0 ......(ii) Now, perpendicular distance from the origin to this plane =