I am currently working on a project where i have micro-controller,stepper drivers etc all powered through a single 12v 10a power supply. I was planning to use the same supply to power the raspberry pi Zero W using a 7805 voltage regulator..
7805 can not give you sufficient amount of current. According to the datasheet it can give you maximum of 1.5A(You need a heat sink and cooler for that). You have to use around 4 7805 IC for this operation, and also a filter system. There is also some other risks, if you are powering your sensors or module from pi then there is a chance of over current through your pi which will damage your board. My suggestion is you must go for a better SMPS. Or try a good quality 5V-2.5A adapteor
A generic 7805 in T0-220 package can supply between 1A and 1.5A (and peak at 2A) but all that depends on adequate heatsinking and sufficient input voltage, this gives you 5W of power. The Pi Zwero W can draw around 1.2W so, on paper, should be well within the capability of a 7805.
However most of the power packs I've seen actually put out 5.1 to 5.2v so I'm not so sure a 7805 would quite work, perhaps consider an adjustable linear regulator like the LM317 instead that you can set to 5.1v.
The other issue you might have is the efficiency, you've included the "battery" tag on your question but only mention that your have a 12v 10A source. As a (simplified) example a 7805 driving 5v at 250mA (or 1.25W) from a 12v source needs to drop 7v, this is turned into 1.75W of wasted heat that you need to dissipate. If this matters to you then a buck converter is going to be more efficient, some of the smaller pre-made modules can be quite affordable and have adjustable outputs and current limits but you'll need to keep an eye on the output noise under load and maybe add decoupling capacitors if it's a problem. Or you can build your own if you're so inclined.
