The value of the determinant […] , is

The correct answer is option (2) : $3abc-a^3-b^3-c^3$

We have,

$\begin{bmatrix}b+c & a-b & a\\c+a & b-c & b\\a+b & c-a & c\end{bmatrix}$

$=\begin{bmatrix}a+b+c & -b & a\\b+c+a & -c & b\\c+a+b & -a & c\end{bmatrix}$               [Applying $C_1→C_1+C_3;C_2→-C_2-C_3$]

$=-(a+b+c) \begin{bmatrix}1 & b & a\\1 & c & b\\1 & a & c\end{bmatrix}$

$=-(a+b+c)\begin{bmatrix}1 & b & a\\0 & c-b & b-a\\0 & a-b & c-a\end{bmatrix}$                $\begin{bmatrix} Applying \\ R_2→R_2-R_1, R_3→R_3-R_1\end{bmatrix}$

$=-(a+b +c) (a^2 + b^2 +c^2 -ab-bc -ca)$

$=-(a^3+b^3+c^3-3abc)$