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(C) Let $\Delta = \begin{vmatrix} b+c & a & a \ b & c+a & b \ c & c & a+b \end{vmatrix}$. Applying $C_1 	o C_1 + C_2 + C_3$: $\Delta = \begin{vmatr

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$4abc$

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(C) Let $\Delta = \begin{vmatrix} b+c & a & a \ b & c+a & b \ c & c & a+b \end{vmatrix}$. Applying $C_1 	o C_1 + C_2 + C_3$: $\Delta = \begin{vmatrix} 2(a+b+c) & a & a \ 2(a+b+c) & c+a & b \ 2(a+b+c) & c & a+b \end{vmatrix}$. Taking $2(a+b+c)$ common from $C_1$: $\Delta = 2(a+b+c) \begin{vmatrix} 1 & a & a \ 1 & c+a & b \ 1 & c & a+b \end{vmatrix}$. Applying $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_1$: $\Delta = 2(a+b+c) \begin{vmatrix} 1 & a & a \ 0 & c & b-a \ 0 & c-a & b \end{vmatrix}$. Expanding along $C_1$: $\Delta = 2(a+b+c) [1(c(b) - (b-a)(c-a))]$. $\Delta = 2(a+b+c) [bc - (bc - ab - ac + a^2)]$. $\Delta = 2(a+b+c) [bc - bc + ab + ac - a^2]$. $\Delta = 2(a+b+c) [ab + ac - a^2]$. $\Delta = 2a(a+b+c) (b+c-a)$. Note: The original determinant simplifies to $4abc$ only if the matrix structure is $\begin{vmatrix} b+c & a & a \ b & c+a & b \ c & c & a+b \end{vmatrix}$. The provided solution in the prompt had a typo in the matrix structure. The correct value is $2a(a+b+c)(b+c-a)$.

The value of $\begin{vmatrix} b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{vmatrix}$ is

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