Unsure of transistor to use for LED strip
Your LED strip runs at \$I_{LED} = 32W/12V = 2.7A\$.  This is the current running through the transistor.

The MPF102 is an N-channel JFET that is not meant to drive a lot of current (datasheet specs 20mA at \$V_{GS} = 0V\$).

The MJE3055T and TIP31A are both bipolar power NPN transistors.  You can approximate them as current controlled current sources; their current amplification, \$h_{FE}\$, is less than 100.  This means that your base current \$I_B\$ will have to be more than \$I_{LED}/100 = 27mA\$.  Your rPi GPIOs will probably not be able to run that amount of current (typical is 4mA for GPIOs).

The TIP102 and TIP120 are both NPN Darlington transistors.  Their \$h_{FE}\$ are well over 1000, so you will not have problems driving the transistors.  The high gain comes at a price, though: the collector-emitter voltage \$V_{CE}\$ is at least 0.7V; at your 2.7A it will be just under 1V (see Fig. 6 in the datasheet).  This means that the power dissipated in the transistor will be \$1V \times 2.7A = 2.7W\$.  Looking at the transistor's thermal resistance \$R_{thj-amb}\$, the junction temperature will be at \$2.7W * 62.5^{\circ}C/W = 170^{\circ}C\$ above ambient temperature.  This exceeds the maximum allowed junction temperature of \$T_J = 150^{\circ}C\$.  Without a heat sink, this transistor will not work either.

If you decide on trying either of the NPN transistors (Darlington or not), be sure to use a base current limiting resistor between your GPIO and the transistor base.  To limit the current at 4mA, you would choose \$R_B = (3.3V-0.7)/4mA = 650\Omega\$.  Of course you do not need exactly this resistance.  A 1k&Omega; or 470&Omega; resistor would do the job as well.