Which of the following determinant(s) vanish( | vedclass

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(A) For option $A$: Let $\Delta_1 = \left| \begin{array}{ccc} 1 & bc & bc(b+c) \ 1 & ca & ca(c+a) \ 1 & ab & ab(a+b) \end{array} ight|$. Multiply

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$\left| \begin{array}{ccc} 1 & bc & bc(b+c) \ 1 & ca & ca(c+a) \ 1 & ab & ab(a+b) \end{array} ight|$

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(A) For option $A$: Let $\Delta_1 = \left| \begin{array}{ccc} 1 & bc & bc(b+c) \ 1 & ca & ca(c+a) \ 1 & ab & ab(a+b) \end{array} ight|$. Multiply $C_1$ by $abc$,then $\Delta_1 = \frac{1}{abc} \left| \begin{array}{ccc} a & abc & abc(b+c) \ b & abc & abc(c+a) \ c & abc & abc(a+b) \end{array} ight|$. Taking $abc$ common from $C_2$ and $C_3$,$\Delta_1 = \frac{(abc)^2}{abc} \left| \begin{array}{ccc} a & 1 & b+c \ b & 1 & c+a \ c & 1 & a+b \end{array} ight| = abc \left| \begin{array}{ccc} a & 1 & a+b+c \ b & 1 & a+b+c \ c & 1 & a+b+c \end{array} ight|$. Since $C_1+C_2$ is proportional to $C_3$,$\Delta_1 = 0$.For option $B$: Let $\Delta_2 = \left| \begin{array}{ccc} 1 & ab & \frac{a+b}{ab} \ 1 & bc & \frac{b+c}{bc} \ 1 & ca & \frac{c+a}{ca} \end{array} ight|$. This determinant does not necessarily vanish for all $a, b, c$.For option $C$: Let $\Delta_3 = \left| \begin{array}{ccc} 0 & a-b & a-c \ b-a & 0 & b-c \ c-a & c-b & 0 \end{array} ight|$. This is a skew-symmetric determinant of odd order $3 	imes 3$. The value of a skew-symmetric determinant of odd order is always $0$. Thus,$\Delta_3 = 0$.

Which of the following determinant$(s)$ vanish(es)?

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