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Consider the lines L(1):(x-1)/(2)=(y)/...

Consider the lines
`L_(1):(x-1)/(2)=(y)/(-1)=(z+3)/(1),L_(2):(x-4)/(1)=(y+3)/(1)=(z+3)/(2)` and the planes `P_(1):7x+y+2z=3," "P_(2):3x+5y-6z=4.` Let `ax+by+cz=d` the equation of the plane passing through the point of intersection of lines` L_(1)` and `L_(2)` and perpendicualr to planes `P_(1)` and `P_(2)`. Match List I with List II and select the correct answer using the code given below the lists.

A

`{:(P,Q,R,S),(3,2,4,1):}`

B

`{:(P,Q,R,S),(1,3,4,2):}`

C

`{:(P,Q,R,S),(3,2,1,4):}`

D

`{:(P,Q,R,S),(2,4,1,3):}`

Text Solution

Verified by Experts

The correct Answer is:
A
`L_(1):(x-1)/(2)=(y-0)/(-1)=(z-(-3))/(1)`
Normal of plane `P:n=|{:(hati,hatj,hatk),(7,1,2),(3,5,-6) :}|`
`=hati(-16)-hatj(-42-6)+hatk(32)`
`=-16hati+48hatj+32hatk`
DR's of normal `n=hati-3hatj-2hatk`
Point of intersection (5,-2,-1)
Now, equation of plane,
`1.(x-5)-3(y+2)-2(z+1)=0`
`implies" "x-3y-2z-13=0`
`implies" "x-3y-2z=13`
`:." "ato1,bto-3,cto-2,d to13`
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