    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 distance2√3from 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=122

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Solution

## The correct option is A 5x−11y+z=17 (i) equation of a plane through intersection of two planes (a1x+b1y+c1z+d1)+λ(a2x+b2y+c2z+d2)=0 (ii) distance of a point (x1,y1,z1) from ax+by+cz+d=0|ax1+by1+cz1+d|√a2+b2+c2 Equation of plane passing through intersection of two planes x+2y+3z=2 and x−y+z=3 is (x+2y+3z−2)+λ(x−y+z−3)=0⇒(1+λ)x+(2−λ)y+(3+λ)z−(2+3λ)=0 Whose distance from (3,1,-1) is 2√3. ⇒|3(1+λ)+1.(2−λ)−1(3+λ)−(2+3λ)|√(1+λ)2+(2−λ)2+(3+λ)2=2√3⇒|−2λ|√3λ2+4λ+14=2√3⇒3λ2=3λ2+4λ+14⇒λ=−72∴(1−72)x+(2+72)y+(3−72)z−(2−212)=0 ⇒−5x2+112y−12z+172=0 or 5x−11y+z−17=0  Suggest Corrections  0      Similar questions  Related Videos   Applications of Second Degree Equations
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