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(D) Let $\Delta = \left|\begin{array}{ccc}a^{2} & b c & c^{2}+a c \ a^{2}+a b & b^{2} & c a \ a b & b^{2}+b c & c^{2}\end{array}ight|$.Taking $a$

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$4$

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(D) Let $\Delta = \left|\begin{array}{ccc}a^{2} & b c & c^{2}+a c \ a^{2}+a b & b^{2} & c a \ a b & b^{2}+b c & c^{2}\end{array}ight|$.Taking $a$ common from $C_1$, $b$ common from $C_2$, and $c$ common from $C_3$:$\Delta = abc \left|\begin{array}{ccc}a & c & \frac{c^2+ac}{c} \ a+b & b & a \ b & b+c & c\end{array}ight| = abc \left|\begin{array}{ccc}a & c & c+a \ a+b & b & a \ b & b+c & c\end{array}ight|$.Applying $C_1 	o C_1 + C_2 - C_3$:$\Delta = abc \left|\begin{array}{ccc}a+c-(c+a) & c & c+a \ a+b+b-a & b & a \ b+b+c-c & b+c & c\end{array}ight| = abc \left|\begin{array}{ccc}0 & c & c+a \ 2b & b & a \ 2b & b+c & c\end{array}ight|$.Applying $R_3 	o R_3 - R_2$:$\Delta = abc \left|\begin{array}{ccc}0 & c & c+a \ 2b & b & a \ 0 & c & c-a\end{array}ight|$.Expanding along $C_1$:$\Delta = abc \cdot (-2b) \cdot \left|\begin{array}{cc}c & c+a \ c & c-a\end{array}ight| = -2ab^2c \cdot (c(c-a) - c(c+a)) = -2ab^2c \cdot (c^2 - ac - c^2 - ac) = -2ab^2c \cdot (-2ac) = 4a^2b^2c^2$.Comparing with $k a^2b^2c^2$, we get $k = 4$.

If $\left|\begin{array}{ccc}a^{2} & b c & c^{2}+a c \\ a^{2}+a b & b^{2} & c a \\ a b & b^{2}+b c & c^{2}\end{array}\right|=k a^{2} b^{2} c^{2}$, then $k=$

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