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Question

# The equation of a plane passing through the line of intersection of the planes x+2y+3z=2 and x−y+z=3 and at a distance 2√3 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 5x−11y+z=17Equation of plane passing through intersection of planes P1 and P2 is P1+λP2=0 (x+2y+3z−2)+λ(x−y+z−3)=0 (1+λ)x+(2−λ)y+(3+λ)z−2−3λ=0 Distance of plane from point (3,1,−1) is 2√3 ⇒ 2√3=|3+3λ+2−λ−3−λ−2−3λ|√(1+λ)2+(2−λ)2+(3+λ)2 ⇒ ∣∣∣−2λ√3λ2+4λ+14∣∣∣=2√3 ⇒ 3λ2=3λ2+4λ+14 ∴ λ=−72 So equation of plane is (1−72)x+(2+72)y+(3−72)z−2+212=0 ∴5x−11y+z=17

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