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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 distance23from 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

5x11y+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 xy+z=3 is
(x+2y+3z2)+λ(xy+z3)=0(1+λ)x+(2λ)y+(3+λ)z(2+3λ)=0
Whose distance from (3,1,-1) is 23.
|3(1+λ)+1.(2λ)1(3+λ)(2+3λ)|(1+λ)2+(2λ)2+(3+λ)2=23|2λ|3λ2+4λ+14=233λ2=3λ2+4λ+14λ=72(172)x+(2+72)y+(372)z(2212)=0
5x2+112y12z+172=0
or 5x11y+z17=0


Mathematics

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