AP EAMCET Engineering 22 Apr 2019 Shift 1 Paper

Consider the diagram,The equation of lines passing through (1,-1,1) and havingdirection cosines as (2,3,1,) are

x−1

2

=

y+1

3

=

z−1

1

=r
From the above equations, we get
x=2r+1,y=3r−1 ,z=r+1
The points, (2r+1,3r−1,r+1) lie on plane.
These points satisfy the equation of the plane,
3(2r+1)+4(3r−1)+5(r+1) +19=0
6r+3+12r−4+5r +5+19=0
23r+23=0
r=−1
So, the point is,
(−2+1,−3−1,−1+1) =(−1,−4,0)
Now the required distance is,
√(−2)2+(−3)2+12 =√4+9+1
=√14