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The intersection of the spheres $x^2+y^2+z^2+7 x-2 y-z=13$ and $x^2+y^2+z^2-3 x+3 y+4 z=8$ is the same as the intersection of one of the sphere and the plane
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Verified Answer
The correct answer is:
$2 x-y-z=1$
$2 x-y-z=1$
Required plane is $S_1-S_2=0$ where $S_1=x^2+y^2+z^2+7 x-2 y-z-13=0$ and
$$
\begin{aligned}
& S_2=x^2+y^2+z^2-3 x+3 y+4 z-8=0 \\
& \Rightarrow 2 x-y-z=1
\end{aligned}
$$
$$
\begin{aligned}
& S_2=x^2+y^2+z^2-3 x+3 y+4 z-8=0 \\
& \Rightarrow 2 x-y-z=1
\end{aligned}
$$
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