What will be the output of the following pseudocode?Integer a,b,cset a= 0, b= 5, c=12if (3<c)a= 10 & cend ifif ((c - a + b) > (b - c))a= ( a + 6 ) + cc = ( a = a ) & bend ifprint a + b + c[ Note- & bitwise AND - The bitwise AND operator ( & ) compares each bit of the first operand to the corresponding bit of the second operand if both bits are 1 the corresponding result bit is set to 1. otherwise, the corresponding result bit is set to 0 ]

The pseudocode provided is a bit confusing due to the use of the assignment operator (=) in some unusual places, but I'll try to interpret it as best as I can.

1. We start with a=0, b=5, c=12.
2. The first if condition checks if 3 is less than c (which is 12), so it's true. Then it performs a bitwise AND operation between 10 and c (12). The result of 10 & 12 is 8 (in binary: 1010 & 1100 = 1000), so now a=8.
3. The second if condition checks if (c - a + b) is greater than (b - c). Substituting the current values, we get (12 - 8 + 5) > (5 - 12), which simplifies to 9 > -7. This is true, so we enter the if block.
4. Inside this block, a is updated to (a + 6) + c, which is (8 + 6) + 12 = 26. Then, c is updated to the result of a bitwise AND operation between a and b. However, the expression (a = a) seems to be a mistake, as it doesn't change the value of a and doesn't make sense in this context. Assuming it's meant to be just 'a', the result of 26 & 5 (in binary: 11010 & 101 = 0), so c=0.
5. Finally, we print a + b + c, which is 26 + 5 + 0 = 31.

So, the output of the pseudocode should be 31, assuming my interpretation of the confusing parts is correct.