Which of the following values of $a$ satisfy the equation $\begin{vmatrix} (1+a)^2 & (1+2a)^2 & (1+3a)^2 \\ (2+a)^2 & (2+2a)^2 & (2+3a)^2 \\ (3+a)^2 & (3+2a)^2 & (3+3a)^2 \end{vmatrix} = -684a$
Step-by-Step Solution
Key Concept: Recognize that the determinant has a special structure where each entry is $(i+ja)^2$ with $i\in\{1,2,3\}$ and $j\in\{1,2,3\}$. Apply row operations $R_2 - R_1$ and $R_3 - R_2$ to exploit the difference of squares formula $(p+q)^2-(p)^2 = 2pq+q^2$, which factors the determinant into linear factors in $a$.
For the determinant with entries $(1+a)^2, (1+2a)^2, (1+3a)^2$ in the first row and similar patterns for subsequent rows, perform row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_2$ to reduce the matrix. This yields a determinant involving differences of squares that simplifies to $-648a$. Further reduction and factoring gives $2(12a^3 - 16a^2) = -648a$, leading to $8a^3 = 648a$, so $a(a^2 - 81) = 0$, giving $a = 0, 9, -9$.
Correct Answer: 2,3