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(C) Let $\Delta = \left| \begin{array}{ccc} 1 & 1 & 1 \ bc & ca & ab \ b+c & c+a & a+b \end{array} ight|$.Applying the column operations $C_1 	o

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

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(C) Let $\Delta = \left| \begin{array}{ccc} 1 & 1 & 1 \ bc & ca & ab \ b+c & c+a & a+b \end{array} ight|$.Applying the column operations $C_1 	o C_1 - C_2$ and $C_2 	o C_2 - C_3$:$\Delta = \left| \begin{array}{ccc} 0 & 0 & 1 \ bc-ca & ca-ab & ab \ (b+c)-(c+a) & (c+a)-(a+b) & a+b \end{array} ight|$$\Delta = \left| \begin{array}{ccc} 0 & 0 & 1 \ -c(a-b) & -a(b-c) & ab \ b-a & c-b & a+b \end{array} ight|$Taking $(b-a)$ common from $C_1$ and $(c-b)$ common from $C_2$:$\Delta = (b-a)(c-b) \left| \begin{array}{ccc} 0 & 0 & 1 \ -c & -a & ab \ 1 & 1 & a+b \end{array} ight|$Expanding along the first row:$\Delta = (b-a)(c-b) [1 \cdot (-c - (-a))] = (b-a)(c-b)(a-c)$$= (-(a-b))(-(b-c))(-(c-a)) = (a-b)(b-c)(c-a)$.

The value of $\left| \begin{array}{ccc} 1 & 1 & 1 \\ bc & ca & ab \\ b+c & c+a & a+b \end{array} \right|$ is

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