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(D) Given the determinant equation: $\left| \begin{array}{ccc} 1+a & 1 & 1 \ 1 & 1+b & 1 \ 1 & 1 & 1+c \end{array} ight| = 0$. Divide $R_1$ by $a$

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(D) Given the determinant equation: $\left| \begin{array}{ccc} 1+a & 1 & 1 \ 1 & 1+b & 1 \ 1 & 1 & 1+c \end{array} ight| = 0$. Divide $R_1$ by $a$,$R_2$ by $b$,and $R_3$ by $c$: $abc \left| \begin{array}{ccc} \frac{1}{a}+1 & \frac{1}{a} & \frac{1}{a} \ \frac{1}{b} & \frac{1}{b}+1 & \frac{1}{b} \ \frac{1}{c} & \frac{1}{c} & \frac{1}{c}+1 \end{array} ight| = 0$. Apply $R_1 ightarrow R_1 + R_2 + R_3$: $abc \left| \begin{array}{ccc} 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} & 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} & 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \ \frac{1}{b} & \frac{1}{b}+1 & \frac{1}{b} \ \frac{1}{c} & \frac{1}{c} & \frac{1}{c}+1 \end{array} ight| = 0$. Taking $(1 + a^{-1} + b^{-1} + c^{-1})$ common from $R_1$: $abc(1 + a^{-1} + b^{-1} + c^{-1}) \left| \begin{array}{ccc} 1 & 1 & 1 \ \frac{1}{b} & \frac{1}{b}+1 & \frac{1}{b} \ \frac{1}{c} & \frac{1}{c} & \frac{1}{c}+1 \end{array} ight| = 0$. Since $a, b, c 
eq 0$,the determinant part simplifies to $1$,thus $1 + a^{-1} + b^{-1} + c^{-1} = 0$. Therefore,$a^{-1} + b^{-1} + c^{-1} = -1$.

If $a, b, c$ are all different from zero and $\left| \begin{array}{ccc} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{array} \right| = 0$,then the value of $a^{-1} + b^{-1} + c^{-1}$ is

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