For any a, b, c in R,the determinant left|beg | vedclass

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(C) Let $\Delta = \left|\begin{array}{lll}bc & b+c & 1 \ ca & c+a & 1 \ ab & a+b & 1\end{array}ight|$.Applying $R_2 ightarrow R_2 - R_1$ and $R_

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$(a-b)(b-c)(c-a)$

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(C) Let $\Delta = \left|\begin{array}{lll}bc & b+c & 1 \ ca & c+a & 1 \ ab & a+b & 1\end{array}ight|$.Applying $R_2 ightarrow R_2 - R_1$ and $R_3 ightarrow R_3 - R_1$:$\Delta = \left|\begin{array}{ccc}bc & b+c & 1 \ c(a-b) & a-b & 0 \ b(a-c) & a-c & 0\end{array}ight|$.Taking $(a-b)$ common from $R_2$ and $(a-c)$ common from $R_3$:$\Delta = (a-b)(a-c) \left|\begin{array}{ccc}bc & b+c & 1 \ c & 1 & 0 \ b & 1 & 0\end{array}ight|$.Applying $R_3 ightarrow R_3 - R_2$:$\Delta = (a-b)(a-c) \left|\begin{array}{ccc}bc & b+c & 1 \ c & 1 & 0 \ b-c & 0 & 0\end{array}ight|$.Expanding along the third column:$\Delta = (a-b)(a-c) \cdot 1 \cdot [0 - (b-c)] = (a-b)(a-c)(-(b-c)) = (a-b)(b-c)(c-a)$.Thus,the correct option is $(c)$.

For any $a, b, c \in R$,the determinant $\left|\begin{array}{lll}bc & b+c & 1 \\ ca & c+a & 1 \\ ab & a+b & 1\end{array}\right|$ is equal to

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