IIT JEE Advanced 2007 Paper 2

Let

x−x1

a

=

y−y1

b

=

z−z1

c

be the equation of line of intersection of plans:
3x - 6y - 2z = 15
2x + y - 2z = 5
Since the line of intersection lies in both planes, we get
3a - 6 - 2c = 0
2a + b - 2c = 0
That is,

a

14

=

b

2

=

c

15

⇒ a = 14 ; b = 2 ; c = 15
Let (x1,y1,0) be a point that lies on the line of intersection; therefore,
3x1−6y1 = 15
2x1+y1 = 5
⇒ (x1,y1,z1) = (3 , - 1 , 0)
Hence, the equation of line of intersection is

x−3

14

=

y+1

2

=

z

15

= t
That is,
x = 3 + 14t
y = 2t - 1
z = 15 t
Therefore, Statement-1 is false and Statement-2 is true.