How can I get a small resolution for the binomial fourfold   $x_1 x_2 x_3- y_1 y_2=0$ in $\mathbb{C}^5$? I consider the singular fourfold $X$ defined as follows:
$$X: \quad x_1 x_2 x_3 -y_1  y_2=0\quad \text{in}\quad \mathbb{C}^5.$$ 
Its  singular locus is a bouquet of three planes meeting at the origin:
$$Sing(X):\quad y_1=y_2=x_1 x_2=x_2 x_3=x_1 x_3=0.$$ 
How can I described the small resolution of this space (if any)?
 A: As Alex Woo says, this is a toric example, and hence can be solved with toric methods. Your variety is $\mathrm{Spec} \ \mathbb{C}[S]$ where $S$ is the semigroup ring generated by $(1,0,0,1)$, $(0,1,0,1)$, $(0,0,1,1)$, $(0,0,0,1)$ and $(1,1,1,2)$. (These correspond to the variables $x_1$, $x_2$, $x_3$, $y_1$ and $y_2$ respectively.) This is a saturated semigroup, so toric methods apply with no subtleties. The dual cone is generated by $(1,0,0,0)$, $(0,1,0,0)$, $(0,0,1,0)$, $(-1,-1,0,1)$, $(-1,0,-1,1)$, $(0,-1,-1,1)$. Toric resolutions of this singularity correspond to triangulations of this cone.
Notice that the six generators of the dual cone all lie in the plane $w+x+y+3z=1$. In this plane, they form a triangular prism. We can draw our pictures in three coordinates by discarding the final coordinate. However, we need to recognize that, if we do this, a lattice point means a point $(x,y,z)$ such that $x+y+z \equiv 1 \mod 3$. (This is the point I screwed up earlier.) 
So we want to understand triangulations of the prism with vertices $(1,0,0)$, $(0,1,0)$, $(0,0,1)$, $(0,-1,-1)$, $(-1,0,-1)$, $(-1,-1,0)$. Now, suppose that our subdivision has a face $d$ which is contained in a face of the triangular prism of dimension $e$. The cones on these faces have dimension $d+1$ and $e+1$; the corresponding torus orbits have dimension $3-d$ and $3-e$, so the fibers here have dimension $e-d$. So smallness means that $2(e-d) < e+1$. For $e=0$, $1$, $2$, $3$ this gives $d \geq 0$, $1$, $1$ and $2$, respectively. In other words, we must add no new vertices to the triangular prism, and we may only add new edges within $2$-faces.
Fortunately, the standard triangulation of the triangular prism has this property. 
There are three tetrahedra:
$$\mathrm{Hull} {\Large (} (1,0,0), \ (0,1,0), \ (0,0,1), \ (0,-1,-1) {\Large )}$$
$$\mathrm{Hull} {\Large (} (0,1,0), \ (0,0,1), \ (0,-1,-1), \ (-1,0,-1) {\Large )}$$
$$\mathrm{Hull} {\Large (} (0,0,1), \ (0,-1,-1), \ (-1,0,-1), \ (-1,-1,0) {\Large )}$$
In my previous update, I worried that these are not unimodular, because the "lattice point" $(0,0,0)$ lay on the $2$-faces of some of them. However, that is actually not a lattice point. (It corresponds to $(0,0,0,1/3)$ back in $4$-space.) Sorry about the confusion.
A: Warning: This seems to be a really bad way of answering this question (but it at least tells you there is one).
The intersection of the opposite Schubert cell $X_\circ^{13425}$ with the Schubert variety $X_{34512}$ is defined by that equation in the appropriate coordinates.  This tells you that locally around the Schubert point $e_{13425}$, the Schubert variety is isomorphic to the product of that fourfold with $\mathbb{C}^{4}$.
Since the permutation $34512$ is 321 and hexagon avoiding, the Bott--Samelson resolution is small for that Schubert variety.
As you have a binomial, I would expect there to be some toric answer that is at least a little more general than this one.
