Given determinant
$\Delta=\left|\begin{array}{ccc}n^2 & (n+1)^2 & (n+2)^2 \\ (n+1)^2 & (n+2)^2 & (n+3)^2 \\ (n+2)^2 & (n+3)^2 & (n+4)^2\end{array}\right|$
On applying $R_2 \rightarrow R_2-R_1$ and $R_3 \rightarrow R_3-R_1$, we get
$\Delta=\left|\begin{array}{ccc}n^2 & (n+1)^2 & (n+2)^2 \\ 2 n+1 & 2 n+3 & 2 n+5 \\ 4 n+4 & 4 n+8 & 4 n+12\end{array}\right|$
On applying $R_3 \rightarrow R_3-2 R_2$, we get
$\Delta=\left|\begin{array}{ccc}n^2 & (n+1)^2 & (n+2)^2 \\ 2 n+1 & 2 n+3 & 2 n+5 \\ 2 & 2 & 2\end{array}\right|$
$\begin{aligned} & n^2(4 n+6-4 n-10)-(n+1)^2 \\ & (4 n+2-4 n-10)+(n+2)^2(4 n+2-4 n-6) \\ & =-4 n^2+8(n+1)^2-4(n+2)^2=-8\end{aligned}$
and it is also given that
$\left|\begin{array}{ccr}1 & -4 & 7 \\ -2 & 3 & -5 \\ 3 & x & -3\end{array}\right|=2 \Delta+1$
$\begin{array}{ll}\Rightarrow & 1(-9+5 x)+4(6+15)+7(-2 x-9) \\ \Rightarrow & 5 x-9+84-14 x-63=-15 \\ \Rightarrow & -9 x+12=-15 \\ \Rightarrow & 9 x=27 \\ & x=3\end{array}$
Hence, option (a) is correct.