Locus of centroid A tangent plane to the ellipsoid $$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1 $$ meets the coordinate axes in A, B, C. 
What will be the locus of the centroid of the triangle ABC?
Please explain the approach in as simple manner as possible.
 A: Any point on the unit sphere can be described using latitude $\lambda$ and longitude $\beta$ as
$$(\cos\beta\sin\lambda,\cos\beta\cos\lambda,\sin\beta)$$
I'd prefer polynomials over trigonometric functions, so I'll apply the tangent half-angle substitution to this, using $t$ and $u$ as parameters.
\begin{align*}
\cos\beta&=\frac{1-t^2}{1+t^2} & \sin\beta&=\frac{2t}{1+t^2} \\
\cos\lambda&=\frac{1-u^2}{1+u^2} & \sin\lambda&=\frac{2u}{1+u^2}
\end{align*}
So with a few exceptions (which correspond to $t=\infty$ or $u=\infty$), any point on the sphere can be described as
$$\left(
\frac{1-t^2}{1+t^2}\frac{2u}{1+u^2},
\frac{1-t^2}{1+t^2}\frac{1-u^2}{1+u^2},
\frac{2t}{1+t^2}
\right)$$
Which means that a point on your axis-aligned ellipsoid can be described as
$$\frac{1}{(1+t^2)(1+u^2)}\left(
a(1-t^2)(2u),
b(1-t^2)(1-u^2),
c(2t)(1+u^2)
\right)$$
Coming from a background of projective geometry, I'd write this as a homogeneous coordinate vector:
$$P=\begin{pmatrix}
a(1-t^2)(2u) \\
b(1-t^2)(1-u^2) \\
c(2t)(1+u^2) \\
(1+t^2)(1+u^2)
\end{pmatrix}$$
The ellipsoid I'd write as a matrix, in the case of the axis-aligned ellipsoid as a diagonal matrix
$$E=\begin{pmatrix}b^2c^2&&&\\&a^2c^2&&\\&&a^2b^2&\\&&&-a^2b^2c^2\end{pmatrix}$$
A point $P$ lies on the ellipsoid $E$ if $P^TEP=0$, which is the case for the above. But more important in our application here, $EP$ describes the tangent plane $A$ in point $P$:
$$A:
bc(1-t^2)(2u)x + ac(1-t^2)(1-u^2)y + ab(2t)(1+u^2)z = abc(1+t^2)(1+u^2)
$$
Now you can set two of these coordinates to zero to obtain the third. This way you get three corners for your triangle. To get the centroid, you add them and divide by three. Or you do this in homogeneous coordinates, adding them after ensuring that they agree in their last coordinate. The centroid I obtained is
$$C=\begin{pmatrix}
at(1+t^2)(1-u^2)(1+u^2) \\
2bt(1+t^2)u(1+u^2) \\
c(1-t^2)(1+t^2)u(1-u^2) \\
6t(1-t^2)u(1-u^2)
\end{pmatrix}$$
As you can see, the coordinates scale with $a,b,c$ in the $x,y,z$ direction, which makes sense since our whole setup is invariant under affine transformations. We might as well have considered tangents to the unit sphere, and only applied that anisotropic scaling by $a,b,c$ at the very end.
If you want to get rid of the parameters $t$ and $u$, you can formulate three equations for $x,y,z$:
\begin{align*}
6t(1-t^2)u(1-u^2)x &= at(1+t^2)(1-u^2)(1+u^2) \\
6t(1-t^2)u(1-u^2)y &= 2bt(1+t^2)u(1+u^2) \\
6t(1-t^2)u(1-u^2)z &= c(1-t^2)(1+t^2)u(1-u^2)
\end{align*}
Then eliminate $t$ and $u$ from these equations. You'd want to use a computer algebra system for this. The way I formulated things, using resultants to eliminate variables, I came up with a resulting variety which had several components. The most interesting of them is characterized by
$$\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2} = 9$$
I guess that's the locus you want, although I haven't investigated where those other components (e.g. $3z=\pm c$) in my result come from. They are almost certainly spurious factors. You can verify that the point $C$ above satisfies the equation I found:
$$
\frac{a^2}{\bigl(at(1+u^2)(1+t^2)(1-u^2)\bigr)^2}+
\frac{b^2}{\bigl(2bt(1+t^2)u(1+u^2)\bigr)^2}+
\frac{c^2}{\bigl(c(1-t^2)(1+t^2)u(1-u^2)\bigr)^2}=
\frac{9}{\bigl(6t(1-t^2)u(1-u^2)\bigr)^2}
$$
Here is a picture of such a surface, created using Surfer 1.1.0. The axes of the hyperboloid were $a\approx0.75,b\approx0.40,c\approx0.50$, but the exact values have little effect on how this surface looks like.

The depicted surface is the intersection of the specified one with a sphere, so the real surface extends beyond those round cuts. It is easy to imagine how the surface extends towards infinity approaching planes which are tangential to the hyperboloid and parallel to the coordinate planes.
