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A line L(1) with direction ratios -3,2,4...

A line `L_(1)` with direction ratios `-3,2,4` passes through the point A(7,6,2) and a line `L_(2)` with directions ratios 2,1,3 passes through the point B(5,3,4). A line `L_(3)` with direction ratios `2,-2,-1` intersects `L_(1)` and `L_(3)` at C and D, resectively. The equation of the plane parallel to line `L_(1)` and containing line `L_(2)` is equal to

A

`x+3y+4z=30`

B

`x+2y+z=15`

C

`2x-y+z=11`

D

`2x+17y-7z=33`

Text Solution

Verified by Experts

The correct Answer is:
D

Equation of plant parallel to `L_(1)` and containing `L_(2)` is
`a(x-5)+b(y-3)+c(z-4)=0`
`a(x-5)+b(y-3)+c(z-4)=0`
`therefore 2a+b+3c=0` and `-3+2b+4c=0`
`rArr a/-2=b/-17=c/7`
So, required plane is `2x+17y-7z=33`.
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