Using a boost or buck converter for a transient load?
@DanielTork asked why it's better to use a 12 V supply and a down convertor for the 5 V supply than a 5 V supply and a boost convertor to generate 12 V for the door strike. 

By example.....
If you use a readily available 5 V Walwart to power the raspberry Pi there would appear no problems with that. The 'Pi Zero likely draws about 100 mA (depends what your are running and how tight the loops) plus another 100 mA for a WiFi adapter. ...Total 200 mA

To create the 12 V @ 270 mA for the striker you need a DC-DC convertor (with an efficiency of about 80%) and it probably has an operating frequency of 50 -100 kHz.
This means that the average current at 5 V will be about 340 mA. 

Total average current from the 5 V supply will range from 200 mA (just the 'Pi) to 540 mA ('Pi and striker).
However, the DC-DC convertor does not draw a nice average current. Because of the duty cycle involved and related to the frequency it will draw current pulses that may be 2-3 times the average current. So when the striker is activated the current pulses drawn from the 5 V supply may approach an Amp. 

To help ameliorate the current pulses you can increase the capacitance on the output of the 5 V supply, and it will certainly help. But you can expect that the 5 V supply will dip. If you are an experienced Raspberry Pi user, you will already know that it's very easy to drop the 'Pi into it's reset zone (about 4.8 V). You would probably use a 5 V  1 - 1.5 A supply in this case, hoping to keep the 5 V constant under all load conditions. 
IMO bad Ju-Ju.

Now let's configure the supplies the other way; and buy a 12 V supply that will provide the required current for the project.
What the average current is depends on how you generate the 5 V for the Raspberry Pi. 


Use a linear regulator for the 5 V. So the average current at 12 volts is 270 + 200 = 470 mA. So the 12 V supply is probably a 700 - 1000 mA rating.
Now if the striker is turned on (a sudden load) the 12 V may dip (maybe by as much as 0.5 - 1 V). However the linear regulator has a drop out voltage probably down at 7 V, so it's output won't be affected at all by the 12 V load dip.
You do have to deal with the dissipation in the linear regulator. It'll be about 1.4 Watts. 
You could use a switching regulator for the 5 V. The average current now will be less. The 200 mA required for the 'Pi results in only about 80 - 100 mA average from the 12 V supply. This means the 12 V supply only needs to supply on average about 270 + 100 = 400 mA. 
In all probability you'd get a 700 - 1000 mA power supply again, but you could perhaps get away with a 500 mA unit since the striker is very intermittent operation.
Power dissipation in the 12 - 5 Buck convertor is very low, perhaps only 70 mW or so.   


So there is the choices to be made.
Either solution could be ok, but I'd just feel much more confident using the 12 V supply.