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Question

The equation of a plane passing through the line of intersection of the planesx+2y+3z=2,x-y+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


Explanation for the correct option

Finding the equation of the plane

Equation for the required plane is :

(x+2y+3z-2)+λ(x-y+z-3)=0(1+λ)x+(2-λ)y+(3+λ)z-(2+3λ)=0

It's distance from (3,1,-1) is 23.So,

23=3(1+λ)+(2-λ)-(3+λ)-(2+3λ)(λ+1)2+(2-λ)2+(3+λ)243=-2λ23λ2+4λ+143λ2+4λ+14=3λ2λ=-72Now,substituevalueofλ-52x+112y-z2+172=0-5x+11y-z+17=05x-11y+z=17

Therefore, the equation of the plane is 5x11y+z=17.

Hence, the correct answer is Option (A).


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