The intersection of the sphere $x^{2} + y^{2} + z^{2} + 7 x - 2 y - z = 13$ and
$x^{2} + y^{2} + z^{2} - 3 x + 3 y + 4_{Z} = 8$ is

x - 2 y - 2 z = 1

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x - 2 y - z = 1

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x - y - z = 1

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2 x - y - z = 1

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Solution

The correct option is D $2 x - y - z = 1$

Let $(a, b, c)$ belongs to the intersection of two spheres.
Then we have $a^{2} + b^{2} + c^{2} + 7 a - 2 b - c - 13 = 0 = a^{2} + b^{2} + c^{2} - 3 a + 3 b + 4 c - 8$
$\Rightarrow 2 a - b - c = 1$ .
Hence any point on the intersection satisfy the equation $2 x - y - z = 1$ .

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x^{2} + y^{2} + z^{2} + 7 x - 2 y - z = 13

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= 13

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