Please be careful with the 220V, will you?

Some math:

Each led strip needs 12V an consumes 72 watts of power. That is:

P = V x I => I = P / V = 72 / 12 => I=6 Amperes each

101 means 100 Ohms, 200 for 2 resistors in series, so each led row can draw max I = V/R = 12/200 = 0.06 Amperes. So there are indeed 6/0.06 = 100 rows of 3 LEDs each, a total of 300 LEDs.

Now that we have the volumes, we can do some more calculations. By adding circuits in parallel, you maintain the supply stable but the total current is the sum of each circuit's current. So, if you add 2 of these strips in series, you will need 12 volts to power them, but 2 x 6 = 12 amperes.

Connecting circuits in series, you maintain the current stable but the total voltage is the sum of the 2 circuits. If you connect for example 2 of these in series, the total current will maintain 6 amperes, but you will need to provide 24 volts instead

(for the previous cases, the connected circuits must have the same voltage and ampere characteristics).

So, if you connect 20 circuits in series, you will need 6 amperes and 240 volts.

Here is a case: You have 2 strips of 5 meters that you can cut at any length (almost). At 5 meters they draw 6 amperes, so it is normal that each 0.5 meters they dray 1/10 of the current, that is 0.6 amperes. So you could cut them every 0.5 meters and then connect them in series. You will have 20 strips connected in series to have 5 meters of strip. So if you rectify the 220 volts with a proper bridge rectifier and some high voltage electrolytic capacitors, you will only need one limiting resistor to control them.

Now, regarding the dimming. LEDs have the advantage that they can be dimmed directly by injecting PWM pulses. Therefore, a high voltage power mosfet such as the IXFH12N50F can be used to directly control the LEDs in series. An optocoupler can be used to couple the gate of the fet with the microcontroller. The power dissipation from this huge 140 Watts circuit will be as less as the I2R losses on the mosfet, that is

P = I x I x Rmosfet = 0.6 x 0.6 x 0.4 = 0.14 Watts. You won't even need a heatsink for this.

Again, please be careful with the 240V