Why an asymmetric geometry with sp3d and sp3d3 hybridization? A molecule in which the central atom is $sp^3d^2$ hybridized ($\ce{SF_6}$), for example, acquires an octahedral shape, which can be explained by assuming that the hybrid orbitals arrange themselves to minimize repulsion amongst themselves. An octahedron makes sense, it the only way to symmetrically arrange six hybrid orbitals on an atom. The case for an $sp^3$ hybridized atom is similar which from a similar logic can be explained to have a tetrahedral geometry. 

But consider the case of $sp^3d$ hybrid orbitals which arrange themselves in a trigonal bypyramidal shape ($\ce{PCl5}$). This is a gross breach of the usual adherence to symmetry. Two bonds are at the axial location and three at the equatorial location. In fact, the axial bonds are slightly longer than the equatorial ones. The case for $sp^3d^3$ hybridization is similar ($\ce{IF7}$) with five orbitals crammed into the equatorial plane and two sticking out above and below it. Why are these arrangements preferred in favor of one with a simple symmetric arrangement of orbitals (which, I think, would minimize the repulsion)?
As already stated in the comments, there is a considerable debate as to whether d-orbitals can be considered to take part in main group elements’ hybridisation at all.

d-Orbital hybridisation was invented by Linus Pauling to explain the bonds in sulfate and sulfur trioxide because he didn’t like the idea of separating charges. However, d-orbitals are far removed energetically compared to s- and p-orbitals — also the reason why 4s is filled before 3d. Technically, before a d-orbital hybridisation can be considered, the next shell s-orbital should be hybridised in first, following the path of lowest energy.

Conveniently, all the cases you presented in your question can be adequately explained using only s- and p-orbitals. See the links for $\ce{SF6}$, $\ce{PCl5}$ and $\ce{IF7}$ respectively. Many thanks to ron for providing them in his comments!