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(A) Given the determinant $\Delta = \left|\begin{array}{ccc}n^2 & (n+1)^2 & (n+2)^2 \ (n+1)^2 & (n+2)^2 & (n+3)^2 \ (n+2)^2 & (n+3)^2 & (n+4)^2\end{

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$3$

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(A) Given the determinant $\Delta = \left|\begin{array}{ccc}n^2 & (n+1)^2 & (n+2)^2 \ (n+1)^2 & (n+2)^2 & (n+3)^2 \ (n+2)^2 & (n+3)^2 & (n+4)^2\end{array}ight|$.Applying row operations $R_2 ightarrow R_2 - R_1$ and $R_3 ightarrow R_3 - R_2$, we get:$\Delta = \left|\begin{array}{ccc}n^2 & (n+1)^2 & (n+2)^2 \ 2n+1 & 2n+3 & 2n+5 \ 2 & 2 & 2\end{array}ight|$.Applying $R_2 ightarrow R_2 - R_1$ again, we get:$\Delta = \left|\begin{array}{ccc}n^2 & 2n+1 & 2 \ 2n+1 & 2 & 0 \ 2 & 0 & 0\end{array}ight| = 2(0 - 4) = -8$.Now, given the second determinant equation:$\left|\begin{array}{ccc}1 & -4 & 7 \ -2 & 3 & -5 \ 3 & x & -3\end{array}ight| = 2\Delta + 1 = 2(-8) + 1 = -15$.Expanding the determinant:$1(-9 + 5x) + 4(6 + 15) + 7(-2x - 9) = -15$.$-9 + 5x + 84 - 14x - 63 = -15$.$-9x + 12 = -15$.$-9x = -27$.$x = 3$.

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