Solving a Calc 3 Problem: Finding a Level Surface at (1,-2,0)

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The problem involves finding the level surface for the function f(x,y,z) = xyz + 3 at the point (1,-2,0). The constant k for the level surface is determined to be 3, leading to the equation xyz = 0. This equation represents three planes in the coordinate system, not just the coordinate axes. The discussion highlights the need to identify which specific plane intersects the given point (1,-2,0). The conclusion is that the equation xyz = 0 effectively describes the level surface sought in the problem.
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This is a problem I got from a Stanford class in calc 3:

Let f(x,y,z)=xyz+3. Find an equation of the level surface that passes through the point (1,-2,0).

This is as far as I have gotten:
The constant for the level surface will be k = xyz + 3 = (1)(-2)(0) + 3 = 3.
The equation is thus 3 = xyz + 3, or xyz = 0.
From this, I understand that the level surface will consist of the coordinate axes, but is there any way to parametrize or otherwise explicitly define this? If not, should xyz = 0 be sufficient as an equation of the level curve? Thanks!
 
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Well, from what I can tell, the equation xyz=0 seems to be what their looking for.
 
It doesn't just consist of the coordinate axes (they don't even go through your point), it consists of three planes. Which plane goes through your point?
 

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