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Question

The equation of the plane whose distance from the origin is unity and which passes through the line x+3y+6=0=3xy4z is

A
x+2y+2z3=0
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B
2x+y+2z3=0
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C
2x+y2z+3=0
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D
x2y2z3=0
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Solution

The correct options are
C 2x+y2z+3=0
D x2y2z3=0
The equation of any plane which contains the given line is x+3y+6+λ(3xy4z)=0
i. e., (1+3λ)x+y(3λ)4λz+6=0
If the distance of this plane from the origin is unity, then
6(1+3λ)2+(3λ)2+16λ2=1
(1+3λ)2+(3λ)2+16λ2=36
26λ2=26λ=±1
The equation of the planes are 2x+y2z+3=0 (for λ=1)
and x2y2z3=0 (for λ=1)
Hence, options C and D are correct.

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