If x, y, z are different from zero

$\left|\begin{array}{ccc}1+x & 1 & 1 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z\end{array}\right|=0$

$\Rightarrow\left|\begin{array}{ccc}x & 0 & -z \\ 0 & y & -z \\ 1 & 1 & 1+z\end{array}\right|=0$        [Applying $R_{2} \rightarrow R_{2}-R_{3}$ and $\left.R_{1} \rightarrow R_{1}-R_{3}\right]$

$\Rightarrow x[y(1+z)+z]+1(y z)=0 \quad$ [Expanding along first column]

$\Rightarrow x[y+y z+z]+y z=0$

$\Rightarrow x y+x y z+x z+y z=0$

$\Rightarrow x y+y z+z x=-x y z$

$\Rightarrow \frac{x y}{x y z}+\frac{y z}{x y z}+\frac{z x}{x y z}=-\frac{x y z}{x y z}$

$\Rightarrow \frac{1}{z}+\frac{1}{x}+\frac{1}{y}=-1$

$\Rightarrow x^{-1}+y^{-1}+z^{-1}=-1$

Hence, the correct option is (d).