The intersection of the spheres x2+y2+z2+7x−2y−z=13 and x2+y2+z2−3x+3y+4z=8 same as the intersection of one of the spheres and the plane

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x-y-z=1

x-2y-z=1

x-y-2z=1

2x-y-z=1

answer is D.

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Detailed Solution

The given spheres are x2+y2+z2+7x−2y−z=13--------(i)and x2+y2+z2−3x+3y+4z=8------(ii)Subtracting (ii) from (i), we get 10x−5y−5z−5=0or 2x−y−z=1

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