AP EAMCET Engineering 2011 Solved Paper

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
=

|3×0−4×0+7×0+13|

√9+16+49

=

13

√74