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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 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. ListIListIIPa=113Qb=2−3Rc=31Sd=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 L1:x−12=y−0−1=z−(−3)1 Normat of plane P:n=∣∣ ∣ ∣∣^i ^j ^k7 1 23 5−6∣∣ ∣ ∣∣=^i(−16)−^j(−42−6)+^k(32)=−16^i+48^j+32^kDR′s of normal n=^i−3^j−2^kPoint of intersection of L1 and L2.⇒2K1+1=K2+4and −k1=k2−3⇒k1=2 and k2=1∴ Point of intersection (5,−2,−1)Now equation of plane,1.(x−5)−3(y+2)−2(z+1)=0⇒ x−3y−2z−13=0⇒ x−3y−2z=13∴ a→1,b→−3,c→−2,d→13

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