IIT JEE Advanced 2010 Paper 1

Plane 1: ax + by + cz = 0 contains line

x

2

=

y

3

=

z

4

. Therefore,
2a + 3b + 4c = 0 (1)
Plane 2: a′x + b′y + c′z = 0 is perpendicular to plane containing lines

x

3

=

y

4

=

z

2

and

x

4

=

y

2

=

z

3

.
Hence, 3a′ + 4b′ + 2c′ = 0 and 4a′ + 2b′ + 3c′ = 0.
⇒

a′

12−4

=

b′

89

=

c′

6−16

⇒8a - b -10c = 0 (2)
From Eqs. (1) and (2), we get

a

−30+4

=

b

32+20

=

c

−2−24

⇒ Equation of plane is x − 2y + z = 0