JEE Advanced 2013 Paper 2

L ‌1 :

x−1

2

=

y

−1

=

z+3

1

Normal of plane P :

→

n

= |

→ i	→ j	→ k
7	1	2
3	5	-6

|
         =

→

i

(-16) -

→

j

(-42-6) +

→

k

(32)
         = -16

→

i

+ 48

→

j

+ 32

→

k

          ⇒

→

n

=

→

i

- 3

→

j

-2

→

k

Point of intersection of L‌1 and L‌2
2k‌1 + 1 = k‌2 + 4
- k‌1 = k‌2 - 3
1 = 3k‌2 - 2
k‌2 = 1
Point of intersection (5,-2,-1)
Plane (x - 5) - 3(y + 7) - 2( z + 1) = 0
x - 3y - 2z - 5 - 6 - 2 = 0
x - 3y - 2z = 13
⇒ a = 1, b = 3, c = -2 , d = 13