my mistake even if we plug in simple nos we still get remainder as 1..hence C is correct

Answer is C.

If you don't see that every sum within the brackets yields an ODD integer just try plugging in numbers (it's important however to figure out that ANY odd number divided by 2 will yield a remainder of 1)
Use the most easiest one, hence w = 1, x = 2, y = 3 and z = 4. Plugging in gives 105 (ODD) -> remainder will be 1
To confirm the answer try another set of numbers: w = 2, x = 3, y = 4 and z = 5. Plugging in, again gives an odd number 315 -> remainder 1
At this point you should realize that any combination of consecutive integers in the given equation (w+x)*(x+y)*(y+z) will yield an odd number and therefore always leave us with a remainder of 1 when divided by 2. -> Answer C

As w,x,y,z are consecutive positive integers, they will be no fraction. Let's assume that these numbers are 2,3,4,5 and according to this \([w +x][x + y][y + z]\) summation is = 5*7*9 and the result is = odd.
Now let's take other consecutive numbers, 5,6,7,8 and summation is = 11*13*15 and the result is odd.

If an odd number is going to be divided by 2, the remainder will be 1 always. That's why the answer is (C).