You never get the rated power out of a solar panel. They are rated with a calibrated lamp that shines at 1000Watts over a surface area of 1 square meter. That only happens on perfect days, on the equator, sun overhead, no clouds, no dust, panels exactly perpendicular to the sun.
I routinely get > rated. 6700 ft elevation, excellent sun, fixed tilt. Here are my insolation numbers:
OP's battery: 400Ah * 12V = 4800Wh
1200W Array will produce a given total per day based on the insolation.
Right now, for me, I'm getting 7.13 hours/day, so:
1200W * 7.13 = 8556Wh
Thus, it would take my array about 4800Wh/8556Wh about 0.56 days to charge the battery.
At my location and panel orientation, 0.56 days of solar is about noon, so the battery would be charged around noon starting at sunrise with no loads being used.
As you can see, the answer isn't simple arithmetic.
First, you need to know your insolation for your location, panel orientation and tilt. Links #5 and #6 in my signature.
Then you take your daily Wh production and compute how many days it takes you to charge. If it's a fraction of a day, 0.5 days is around noon in most cases (mine isn't due to my panels facing 165° instead of 180°) and you estimate from there.
If you're using power while you're charging your batteries, it just gets that much uglier. If you run the numbers, then at least you can determine how much power you can use in a day and still get your batteries charged.
In the above example, I have 8556-4800Wh = 3756Wh of power I can use and still have my batteries fully charged at the end of the charging day.