If a, b and 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 $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ is :

-1

$\Delta=\left|\begin{array}{ccc}1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c\end{array}\right|=0 \Rightarrow a b c\left|\begin{array}{ccc}\frac{1}{a}+1 & \frac{1}{a} & \frac{1}{a} \\ \frac{1}{b} & \frac{1}{b}+1 & \frac{1}{b} \\ \frac{1}{c} & \frac{1}{c} & \frac{1}{c}+1\end{array}\right|=0$

So applying

$C_2 \rightarrow C_2-C_1$

$C_3 \rightarrow C_3-C_1$

$a b c\left|\begin{array}{ccc}\frac{1}{a}+1 & -1 & -1 \\ \frac{1}{b} & 1 & 0 \\ \frac{1}{c} & 0 & 1\end{array}\right|=0$

$R_1 \rightarrow R_1+R_2+R_3=a b c\left|\begin{array}{ccc}\frac{1}{a} + \frac{1}{b}+\frac{1}{c}+1 & 0 & 0 \\ \frac{1}{b} & 1 & 0 \\ \frac{1}{c} & 0 & 1\end{array}\right|=0$

expanding across $R_1$

$abc(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} +1)=0$

$\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=-1$