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+4z−9=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 A4x+4y+4z−9=0 Assume S1:x2+y2+z2+2x+2y+2z+2=0 and S2:4x2+4y2+4z2+4x+4y+4z−1=0 Therefore plane of intersection is given by, S2−4S1=0 ⇒(4x2+4y2+4z2+4x+4y+4z−1)−4(x2+y2+z2+2x+2y+2z+2)=0 ⇒4x+4y+4z−9=0