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Question

The equation of a plane passing through the line of intersection of the planes x+2y+3z=2 and xy+z=3 and at a distance 23 from the point (3,1,1) is

A
5x11y+z=17
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B
2x+y=321
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C
x+y+z=3
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D
x2y=12
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Solution

The correct option is A 5x11y+z=17
Equation of plane passing through intersection of planes P1 and P2 is P1+λP2=0
(x+2y+3z2)+λ(xy+z3)=0
(1+λ)x+(2λ)y+(3+λ)z23λ=0

Distance of plane from point (3,1,1) is 23
23=|3+3λ+2λ3λ23λ|(1+λ)2+(2λ)2+(3+λ)2
2λ3λ2+4λ+14=23
3λ2=3λ2+4λ+14
λ=72
So equation of plane is
(172)x+(2+72)y+(372)z2+212=0
5x11y+z=17


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