KEAM 2014 Math Paper

Given equation of planes are
x−2y−z+5=0
and ‌‌x+y+3z=6
Firstly, determine the intersection lines of two planes.
Let the DR's of intersection line are a,b and c
since, the normal to the given planes are perpendicular to the intersecting line.
∴‌‌a(1)+b(−2)+c(−1)=0
⇒‌‌a−2b−c=0....(i)
and ‌‌a(1)+b(1)+c(3)=0
⇒‌‌a+b+3c=0....(ii)
From Eqs. (i) and (ii), we get
‌

a

−6+1

=‌

b

−1−3

=‌

c

1+2

‌

a

−5

=‌

b

−4

=‌

c

3

since, the required line is passing through (2,3,1) and parallel to the line of intersection.
∴‌‌‌

x−2

−5

=‌

y−3

−4

=‌

z−1

3