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.Note that all points in the plane of $latex ABC$ satisfy $latex x+y+z=1$ (why?). For any $latex P$ in the interior, Let $latex Q$ be the foot of the perpendicular from $latex P$ to $latex AB$ and $latex R$ be the foot of the perpendicular from $latex P$ to the XY plane. It is possible to show that $latex \frac{PR}{PQ}= \sqrt{\frac{2}{3}}$ (it is because the angle between the plane and the Z axis is $latex \arccos \sqrt{\frac{2}{3}}$). Hence, $latex \frac{z}{c^2}=\sqrt{\frac{3}{2}}$. By symmetry, $latex \frac{x}{a^2}=\sqrt{\frac{2}{3}}=\frac{y}{b^2}$. This relates the two coordinate systems.For a triangle with sides $latex a,b,c$, the square of the area is$latex \frac{1}{16}(a+b+c)(-a+b+c)(a-b+c)(a+b-c)$. For this to be positive, we must have (after simplification)$latex a^4+b^4+c^4< 2(a^2b^2+b^2c^2+c^2a^2)$. In cartesian coordinates, this translates to$latex \frac{3}{2}(x^2+y^2+z^2)<3(xy+yz+zx)$, which is equivalent to $latex (x+y+z)^2>2(x^2+y^2+z^2)$. As $latex P$ lies on the plane $latex x+y+z=1$, this means that $latex x^2+y^2+z^2<\frac{1}{2}$. This last equation is that of the interior of a solid sphere. Hence, our desired locus is the intersection of this solid sphere with the plane $latex x+y+z=1$, which is precisely the interior of the circumcircle of $latex ABC$. 

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