Introduction to 3DHard
Question
Consider the lines L1 : 
L2 : 
and the planes P1: 7x+y + 2z = 3, P2 : 3x + 5y - 6z = 4. Let ax + by + cz = d be the equation of the plane passing through the point of intersection of lines L1 and L2 and perpendicular to planes P1 and P2.
Match List-I with List-II and select the correct answer using the code given below the lists.
List-I List-II
P. a = 1. 13
Q. b = 2. -3
R. c = 3. 1
S. d = 4. -2
L2 : and the planes P1: 7x+y + 2z = 3, P2 : 3x + 5y - 6z = 4. Let ax + by + cz = d be the equation of the plane passing through the point of intersection of lines L1 and L2 and perpendicular to planes P1 and P2.Match List-I with List-II and select the correct answer using the code given below the lists.
List-I List-II
P. a = 1. 13
Q. b = 2. -3
R. c = 3. 1
S. d = 4. -2
Options
A.P → 3, Q → 2, R → 4, S → 1
B.P → 1, Q → 3, R → 4, S → 2
C.P → 3, Q → 2, R → 1, S → 4
D.P → 2, Q → 4, R → 1, S → 3
Solution
For point of intersection of L1 and L2
point of intersction is (5, -2, -1)
Now, vector normal to the plane is
Let equation of required plane be x - 3y - 2z = α
∴ it passes through (5, -2, -1)
∴ α = 13 ⇒ equation of plane is x - 3y - 2z = 13
point of intersction is (5, -2, -1)Now, vector normal to the plane is
Let equation of required plane be x - 3y - 2z = α
∴ it passes through (5, -2, -1)
∴ α = 13 ⇒ equation of plane is x - 3y - 2z = 13
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