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Question

The plane of intersection of x2+y2+z2+2x+2y+2z+2=0 and 4x2+4y2+4z2+4x+4y+4z−1=0

A
4x+4y+4z9=0
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B
x+y+z+9=0
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C
4x+4y+4z+1=0
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D
x+y+z+1=0
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Solution

The correct option is A 4x+4y+4z9=0
Assume S1:x2+y2+z2+2x+2y+2z+2=0 and S2:4x2+4y2+4z2+4x+4y+4z1=0
Therefore plane of intersection is given by,
S24S1=0
(4x2+4y2+4z2+4x+4y+4z1)4(x2+y2+z2+2x+2y+2z+2)=0
4x+4y+4z9=0

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