Curiosity of the application of groebner bases to problems involving finding the envelope of a family of curves drove me to verify the following:
Given a quadratic Bézier curve with control points $(1,0), (0,0) $ and $(0,1)$, can I find this curve back as as the envelope of the family of lines parametrized by $a$ : $f(x,y,a) = a^2-ax+ay-a+x$. As one can verify for each $a$ the corresponding line passes through $(a,0)$ and $(0,1-a)$. Two nearby lines (for $a$ and $a+\epsilon)$ can be approximated by $f(x,y,a)$ and $f(x,y,a)+\epsilon \frac{\partial f(x,y,a)}{\partial a} + \epsilon^2\ldots$ so in the limit $\epsilon \to 0$ the intersection point belongs to the variety associated with the ideal generated by $f(x,y,a)$ and $\frac{\partial f(x,y,a)}{\partial a}$. Applying the groebnerbasis calculation on $[a^2-ax+ay-a+x, \;2a-x+y-1]$ using lex monomial ordering gives as final polynomial (divided by 2): $ (x-y)^2 -2(x+y) +1$ which seamslessly correspond to this special case of a Bézier curve.
When, driven by curiousity, I tried an analogous example in three dimensions. The result was most disappointing. Just like in the two dimensional case I started off with a family of planes that intecepted the axes on three values $a,b,$ and $c$ with the property that $a+b+c = 1$. This gave me a 2-parameter family of planes with equations $f(a,b,x,y,z) = a^2b-a^2y+ab^2-abx-aby+abz-b^2x-ab+ay+bx$. This combined with the polynomials $\frac{\partial f}{\partial a} = 2ab-2ay+b^2-bx-by+bz-b+y\ $ and $\frac{\partial f}{\partial b} = a^2+2ab-ax-ay+az-2bx-a+x\ $. Calulating the groebnerbasis for these three polynomials, using GAP gave (using lex monomial ordering) 29 polynomials with huge rational coefficients (the numerators and denominators have several dozens of digits). Moreover the last six only contain $x, y,z$ and are all divisible by $xyz$. What does this all mean, I'm completely at a loss here.
EDIT: Thanks to the remark of @ahulpke using "ReducedGroebnerBasis" I obtained the polynomial $$x^4-4x^3y-4x^3z+6x^2y^2+4x^2yz+6x^2z^2-4xy^3+4xy^2z+4xyz^2-4xz^3+y^4-4y^3z+6y^2z^2-4yz^3+z^4-4x^3+4x^2y+4x^2z+4xy^2-40xyz+4xz^2-4y^3+4y^2z+4yz^2-4z^3+6x^2+4xy+4xz+6y^2+4yz+6z^2-4x-4y-4z+1 $$ which is symmetric in $x,y,z$ and so can be expressed in the elementary symmetric polynomials $u=x+y+z, v = xy+yz+zx$ and $w=xyz$ as: $$ u^4+16v^2-8u^2v - 4u^3+16uv-64w + 6u^2-8v-4u+1 $$ but I have no clue what this represents geometrically.