Question

The intersection of the spheres $x^2 + y^2 + z^2 + 7x − 2y − z = 13$ and $x^2 + y^2 + z^2 − 3x + 3y + 4z = 8$ is the same as the intersection of one of the sphere and the plane

• A. x - y - z = 1
• B. x - 2y - z = 1
• C. x - y - 2z = 1
• D. 2x - y - z = 1

Answer 1

Answer: (D)

Required plane is $S_{1} - S_{2} = 0$
where $S_{1} = x_{2} + y_{2} + z_{2} + 7x - 2y - z - 13 = 0$ and
$S_{2} = x_{2} + y_{2} + z_{2} - 3x + 3y + 4z - 8 = 0$
$⇒ 2x - y - z = 1.$