What will be the output of the following pseudo code?1. Integer a, b, c2. Set a=3, b=6, c=63. b = (c^1)+b4. for (each c from 2 to 3)5. a = (b&a) +b6. b = (12 + 12) +a7. End for8. b=(b^c)&c9. Print a+b

Let's break down the pseudo code:

1. Integer a, b, c: This line is declaring three variables a, b, and c.

2. Set a=3, b=6, c=6: This line is initializing the variables a, b, and c with the values 3, 6, and 6 respectively.

3. b = (c^1)+b: This line is performing a bitwise XOR operation on c and 1, and then adding the result to b. The XOR operation will return 1 if the two bits are different and 0 if they are the same. So, c^1 = 6^1 = 7. Then, 7 + b = 7 + 6 = 13. So, b is now 13.

4. for (each c from 2 to 3): This line is starting a loop where c will take on the values 2 and 3.

5. a = (b&a) +b: This line is performing a bitwise AND operation on b and a, and then adding the result to b. The AND operation will return 1 if both bits are 1 and 0 otherwise. So, for c=2, a = (13&3) + 13 = 1 + 13 = 14. For c=3, a = (13&14) + 13 = 12 + 13 = 25.

6. b = (12 + 12) +a: This line is adding 12 and 12, and then adding the result to a. So, for c=2, b = 24 + 14 = 38. For c=3, b = 24 + 25 = 49.

7. End for: This line is ending the loop.

8. b=(b^c)&c: This line is performing a bitwise XOR operation on b and c, and then performing a bitwise AND operation on the result and c. So, b = (49^3)&3 = 50&3 = 2.

9. Print a+b: This line is printing the sum of a and b. So, the output will be 25 + 2 = 27.