Heat and Thermal ExpansionHard

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

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

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