Please explain for part 3 a why is D2=0008 6C9A

Question 1 (6 Points) Assume the following register and memory contents in an MC68000 computer: Registers AO, A, A2 and A3 contain S1000, $3000, $5000, and $7000, respectively. Registers DO, DI, and D2 contain $102000, $44000, and S86000, respectively. The address label LOC represents address $41186. Memory locations S1000, $3000, $41186, S5000, and $7000 contain the long words SFFO SFFOOF, SFFOOFF1, SFF00FFF1, and SFFOOFFFF respectively For each of the following assembly instructions, specify: i) The addressing modes of the two operands being used. ii The effective addresses of the two operands being used (i.e., source operand and destination operand). ii) The effect of executing each of the following three instructions, starting each time from the initial state. Each part is worth 2 points. a, ADDLL #3226,D2 a.i. [0.5 point] Immediate, Data register direct a.ii. [0.25 point] EA (source operand)-Immediate operand, part of the instruction i.e., operand- 3226 in decimal). [0.25 point] EA (destination operand) D2 i.e, operand $86000). a.iii. [I point] Adds 3226 to D2.L and overwrites D2.L by the result D2] 0008 6C9A (i.e. 552090 in decimal)