\(x^2-y^2+z^2=2\)
If \(y=2\), then you get \(x^2-2^2+z^2=2\iff x^2+z^2=6\) which is a circle in the x-z plane with radius \(\sqrt6\).
If \(x=1\), then you get \(1^2-y^2+z^2=2\iff z^2-y^2=1\) which is a hyperbola in the y-z plane.
If \(y=0\), you get another circle in the x-z plane defined by \(x^2+z^2=2\), this time with radius \(\sqrt2\).
If \(x=2\), then \(2^2-y^2+z^2=2\iffy^2-z^2=2\), another hyperbola in the y-z plane with branches perpendicular to the case where \(x=1\).
