In the datasheet
http://www.ti.com/lit/ds/symlink/lm317.pdf , figure 8.3.3 says how to calculate Rsc
Imax = 1.2/Rsc, so Rsc = 1.2/Imax. In your schematic Imax is 1.5A, so Rsc = 1.2V/1.5A = 0.8 ohms. The dissipated power for a 0.8 ohm resistor at 1.5A is P = I*I*R = 1.5A*1.5A*0.8ohm = 1.8W
In conclusion Rsc is 0.8

/ 2W. A 2W resistor at least, 3W or more will be even better.
Later edit:
About the -10V, it's a negative voltage. To power that schematic, 2 voltages will be needed. One is +32...40V, the main power voltage. The other voltage is -10V, for the Q1 and Q2 biasing circuits. Q1 and Q2 are constant current generators, in order to get a constant current on each of their diodes pairs (D1, D2 and D3, D4 respectively) and thus a constant 1.2V on each pair of the series diodes.
Both +32...40V and -10V are with respect to the GND (measured relative to the GrouND).
The positive voltage will need to be able to provide at least 1.5A, while for the negative voltage there is not much need for a lot of current, a 0.1A maximum for the -10V will be more than enough.