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(D) Let $\Delta = \left|\begin{array}{ccc}a & b & c \ a^2 & b^2 & c^2 \ 1 & 1 & 1\end{array}ight|$.Option $(A)$: $\left|\begin{array}{ccc}a+1 & b+

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$\left|\begin{array}{ccc}a+b & b+c & c+a \ a^2+b^2 & b^2+c^2 & c^2+a^2 \ 2 & 2 & 2\end{array}ight|$

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(D) Let $\Delta = \left|\begin{array}{ccc}a & b & c \ a^2 & b^2 & c^2 \ 1 & 1 & 1\end{array}ight|$.Option $(A)$: $\left|\begin{array}{ccc}a+1 & b+1 & c+1 \ a^2+1 & b^2+1 & c^2+1 \ 1 & 1 & 1\end{array}ight| = \left|\begin{array}{ccc}a & b & c \ a^2 & b^2 & c^2 \ 1 & 1 & 1\end{array}ight| + \left|\begin{array}{ccc}1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 1\end{array}ight| = \Delta + 0 = \Delta$.Option $(B)$: Applying $C_1 ightarrow C_1-C_2$ and $C_2 ightarrow C_2-C_3$,we get $\left|\begin{array}{ccc}a-b & b-c & c \ a^2-b^2 & b^2-c^2 & c^2 \ 0 & 0 & 1\end{array}ight|$. Expanding along $R_3$,we get $(a-b)(b^2-c^2) - (b-c)(a^2-b^2) = (a-b)(b-c)(b+c) - (b-c)(a-b)(a+b) = (a-b)(b-c)(b+c-a-b) = (a-b)(b-c)(c-a)$,which is the value of $\Delta$.Option $(C)$: $\left|\begin{array}{ccc}a(a+1) & b(b+1) & c(c+1) \ a+1 & b+1 & c+1 \ -1 & -1 & -1\end{array}ight|$. Adding $R_2$ to $R_1$,we get $\left|\begin{array}{ccc}a^2+2a+1 & b^2+2b+1 & c^2+2c+1 \ a+1 & b+1 & c+1 \ -1 & -1 & -1\end{array}ight| = \left|\begin{array}{ccc}(a+1)^2 & (b+1)^2 & (c+1)^2 \ a+1 & b+1 & c+1 \ -1 & -1 & -1\end{array}ight|$. This simplifies to $\Delta$.Option $(D)$: The determinant $\left|\begin{array}{ccc}a+b & b+c & c+a \ a^2+b^2 & b^2+c^2 & c^2+a^2 \ 2 & 2 & 2\end{array}ight|$ is not equal to $\Delta$.

$\left|\begin{array}{ccc}a & b & c \\ a^2 & b^2 & c^2 \\ 1 & 1 & 1\end{array}\right|$ is not equal to

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