If \(a\), \(b\), and \(c\) are consecutive integers, where \(a < b < c\), we can let:
- \(a = x-1\)
- \(b = x\)
- \(c = x+1\)
Then, we need to determine which of the following cannot be the value of \(c^2 - (a^2 + b^2)\).
First, we express \(c^2 - (a^2 + b^2)\) in terms of \(x\):
\[
c^2 - (a^2 + b^2) = (x+1)^2 - [(x-1)^2 + x^2]
\]
Calculating each term, we get:
\[
(x+1)^2 = x^2 + 2x + 1
\]
\[
(x-1)^2 = x^2 - 2x + 1
\]
\[
x^2 = x^2
\]
Thus:
\[
(x+1)^2 - [(x-1)^2 + x^2] = x^2 + 2x + 1 - [x^2 - 2x + 1 + x^2]
\]
\[
= x^2 + 2x + 1 - (x^2 - 2x + 1 + x^2)
\]
\[
= x^2 + 2x + 1 - x^2 + 2x - 1 - x^2
\]
\[
= -x^2 + 4x
\]
\[
= x(4-x)
\]
We need to determine which of the given values cannot be produced by the expression \(x(4-x)\):
\[
\text{A. } -21 \quad (-3 \cdot 7 = -21)
\]
\[
\text{B. } -12 \quad (-2 \cdot 6 = -12)
\]
\[
\text{C. } -6 \quad (-1 \cdot 5 = -5 \text{, not -6})
\]
\[
\text{D. } 0 \quad (0 \cdot 4 = 0)
\]
\[
\text{E. } 3 \quad (1 \cdot 3 = 3)
\]
The value \(-6\) cannot be the product of \(x(4-x)\) for any integer \(x\).