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Question

# Consider the lines L1 : x−12=y−1=z+31, L2: x−41=y+31=z+32 and the planes P1: 7x+y+2z=3, P2: 3x+5y−6z=4. Let ax+by+cz=d be the equation of plane passing through the point of intersection of the lines L1 and L2, and perpendicular to the planes P1 and P2. Match List - I with the List - II and select the correct answer using the code given below the lists : List IList IIPa=1.13Qb=2.−3Rc=3.1Sd=4.−2

A
(P)(3),(Q)(2),(R)(4),(S)(1)
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B
(P)(1),(Q)(3),(R)(4),(S)(2)
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C
(P)(3),(Q)(2),(R)(1),(S)(4)
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D
(P)(2),(Q)(4),(R)(1),(S)(3)
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Solution

## The correct option is A (P)→(3),(Q)→(2),(R)→(4),(S)→(1)Any point on line L1 (2λ+1,−λ,λ−3) Any point on line L2 (μ+4,μ−3,2μ−3) For point of intersection of L1 and L2 ⇒ 2λ+1=μ+4,−λ=μ−3,λ−3=2μ−3 So λ=2,μ=1 ∴ Point of intersection is (5,−2,−1) Let normal of palne ax+by+cz=d be →n →n=∣∣ ∣ ∣∣^i^j^k71235−6∣∣ ∣ ∣∣ ⇒→n=−16^i+48^j+32^k ∴ →n=−^i+3^j+2^k Hence equation of plane is ⇒ −1(x−5)+3(y+2)+2(z+1)=0 ∴ x−3y−2z=13 So a=1,b=−3, c=−2, d=13

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