If a neq b neq c, Delta_1=left|begin{array}{l | vedclass

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(D) First, we simplify $\Delta_1$:$\Delta_1 = \left|\begin{array}{ccc}1 & a^2 & bc \ 1 & b^2 & ca \ 1 & c^2 & ab\end{array}ight|$Applying $R_2 i

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$6(ab+bc+ca)$

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(D) First, we simplify $\Delta_1$:$\Delta_1 = \left|\begin{array}{ccc}1 & a^2 & bc \ 1 & b^2 & ca \ 1 & c^2 & ab\end{array}ight|$Applying $R_2 ightarrow R_2 - R_1$ and $R_3 ightarrow R_3 - R_1$:$\Delta_1 = \left|\begin{array}{ccc}1 & a^2 & bc \ 0 & b^2-a^2 & c(a-b) \ 0 & c^2-a^2 & b(a-c)\end{array}ight| = (b-a)(c-a) \left|\begin{array}{ccc}1 & a^2 & bc \ 0 & b+a & -c \ 0 & c+a & -b\end{array}ight|$Expanding along the first column: $(b-a)(c-a) [-(b+a)b + c(c+a)] = (b-a)(c-a) [-b^2-ab+c^2+ac] = (b-a)(c-a) [(c-b)(c+b) + a(c-b)] = (b-a)(c-a)(c-b)(a+b+c) = (a-b)(b-c)(c-a)(a+b+c)$.Next, we simplify $\Delta_2$:$\Delta_2 = \left|\begin{array}{ccc}1 & 1 & 1 \ a^2 & b^2 & c^2 \ a^3 & b^3 & c^3\end{array}ight| = abc \left|\begin{array}{ccc}1/a & 1/b & 1/c \ a & b & c \ a^2 & b^2 & c^2\end{array}ight|$ (or using standard properties of Vandermonde-like determinants):$\Delta_2 = (a-b)(b-c)(c-a)(ab+bc+ca)$.Given $\frac{\Delta_1}{\Delta_2} = \frac{6}{11}$, we have $\frac{(a-b)(b-c)(c-a)(a+b+c)}{(a-b)(b-c)(c-a)(ab+bc+ca)} = \frac{a+b+c}{ab+bc+ca} = \frac{6}{11}$.Therefore, $11(a+b+c) = 6(ab+bc+ca)$.

If $a \neq b \neq c$, $\Delta_1=\left|\begin{array}{lll}1 & a^2 & b c \\ 1 & b^2 & c a \\ 1 & c^2 & a b\end{array}\right|$, $\Delta_2=\left|\begin{array}{ccc}1 & 1 & 1 \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3\end{array}\right|$ and $\frac{\Delta_1}{\Delta_2}=\frac{6}{11}$, then $11(a+b+c)=$

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