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Let's evaluate the determinant:
\[\begin{vmatrix}-a&b&c\\ a&-b&c\\ a&b&-c\end{vmatrix}\]Apply \(R_1 \rightarrow R_1 + R_2\):
\[\begin{vmatrix}0&0&2c\\ a&-b&c\\ a&b&-c\end{vmatrix}\]Expand along the first row:
\[0 \cdot C_{11} + 0 \cdot C_{12} + 2c \cdot C_{13} = 2c \begin{vmatrix}a&b\\ a&b\end{vmatrix}\]Oops! There was an error in the row operation. Let's try a different approach.
Original determinant:
\[\begin{vmatrix}-a&b&c\\ a&-b&c\\ a&b&-c\end{vmatrix}\]Apply \(C_1 \rightarrow C_1 + C_2\):
\[\begin{vmatrix}-a+b&b&c\\ a-b&-b&c\\ a+b&b&-c\end{vmatrix}\]This doesn't seem to simplify things much either.
Let's try adding all rows to the first row: \(R_1 \rightarrow R_1 + R_2 + R_3\)
\[\begin{vmatrix}a&b&c\\ a&-b&c\\ a&b&-c\end{vmatrix}\]Applying \(R_1 \rightarrow R_1 + R_2 + R_3\):
\[\begin{vmatrix}a+a+a&b-b+b&c+c-c\\ a&-b&c\\ a&b&-c\end{vmatrix} = \begin{vmatrix}a&b&c\\ a&-b&c\\ a&b&-c\end{vmatrix}\]This is not correct either.
Let's evaluate the determinant directly:
\(\begin{vmatrix}-a&b&c\\ a&-b&c\\ a&b&-c\end{vmatrix} = -a((-b)(-c) - (c)(b)) - b((a)(-c) - (c)(a)) + c((a)(b) - (-b)(a))\)
\(= -a(bc - bc) - b(-ac - ac) + c(ab + ab)\)
\(= -a(0) - b(-2ac) + c(2ab)\)
\(= 0 + 2abc + 2abc\)
\(= 4abc\)
Therefore, \(kabc = 4abc\), which means \(k = 4\).
Correct Answer: 4
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