Since x, y, and z are consecutive multiples of 3, let’s write all three integers in terms of x:

x = x

y = x + 3

z = x + 6

So, for Quantity A, the sum can be rewritten only in terms of x:

(x + 1) + (y – 2) + (z + 3)

(x + 1) + (x + 3 – 2) + (x + 6 + 3) {substituting for y and z}

3x + (1 + 3 – 2 + 6 + 3)

3x + 11

Now we find the remainder when this simplified sum is divided by 9.

(Remainder when (3x + 11) div 9) = (Remainder when 3x div 9) + (Remainder when 11 div 9)

Since x is a multiple of 3, 3x is a multiple of 9, and the remainder when 3x is divided by 9 is 0.

When 11 is divided by 9, 9 goes into 11 once, leaving 11 – 9 = 2 as the remainder.

Thus, Quantity A is 0 + 2 = 2.

The correct answer is C.

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