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

[adartsesirhc]

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!

 

[foxjwill]

Well, from what I can tell, the equation xyz=0 seems to be what their looking for.

 

[Dick]

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?