Searching for resolutions of generalized determinental varieties

I'm interested in studying a certain generalization of determinental varieties as defined here: https://en.wikipedia.org/wiki/Determinantal_variety

To be more specific, I must first lay out a few definitions.
Consider the variety $$X=Mat_{n_1, n_2}(\mathbb{C})\times Mat_{n_2, n_3}(\mathbb{C})\times... Mat_{n_{k-1}, n_k}(\mathbb{C}).$$

There exists an action of $$(g_1, ..., g_k)\in GL_{n_1}(\mathbb{C})\times... \times GL_{n_k}(\mathbb{C})$$ on $$X$$ given by $$(g_1, ..., g_k)\cdot(x_1, x_2, ..., x_{k-1})=(g_1x_1g_2^{-1}, ..., g_{k-1}x_{k-1}g_k^{-1}).$$

Letting $$r_{ij}$$ denote the rank of $$x_ix_{i+1}... x_j$$, the orbits are completely determined by all valid choices for the values of $$r_{ij}$$. One can see from this definition that when $$k=2$$, choosing any $$0\leq r_{12}\leq \min\{n_1, n_2\}$$ precisely gives us the case of determinental varieties of rank $$r_{12}$$.
In this case, it is known that the closure is given by all of those elements of $$M_{n_1, n_2}(\mathbb{C})$$ or rank at most $$r_{12}$$. As indicated on the wiki page, these closures are singular, and have well known resolutions.

I believe that in the more general case I have outlined above, the closures of the orbits will also be singular, where the "=$$r_{ij}$$"'s determining the orbits, will be replaced with all "$$\leq r_{ij}$$" for the closures, and that these closures will, in general, be singular.

Question: I would like to know if these varieties have been studied in general by algebraic geometers, if so, what is their name? Are there known resolutions of the closures, and bonus points if we have a good understanding of their fibers, or rather, the cohomology of the fibers.

Note, that is this paper of Zelevinsky: http://www.math.utah.edu/~ptrapa/math-library/zelevinsky/zelevinski_tworemarks.pdf though dressed up differently, the varieties I'm interested in studying are embedded inside Schubert varieties. I think the implication being that one might know where to go from here?

Thanks.