If $x,y,z are not equal to zero$ and $\left|\begin{array}{ccc}1+\mathrm{x}& 1& 1\\ 1+\mathrm{y}& 1+2\mathrm{y}& 1\\ 1+\mathrm{z}& 1+\mathrm{z}& 1+3\mathrm{z}\end{array}\right|=0$, then $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=$ 

$\left|\begin{array}{ccc}1+\mathrm{x}& 1& 1\\ 1+\mathrm{y}& 1+2\mathrm{y}& 1\\ 1+\mathrm{z}& 1+\mathrm{z}& 3+3\mathrm{z}\end{array}\right|$

                $\begin{array}{c}=\mathrm{xyz}\left|\begin{array}{ccc}1+\frac{1}{\mathrm{x}}& \frac{1}{\mathrm{x}}& \frac{1}{\mathrm{x}}\\ 1+\frac{1}{\mathrm{y}}& 2+\frac{1}{\mathrm{y}}& \frac{1}{\mathrm{y}}\\ 1+\frac{1}{\mathrm{z}}& 1+\frac{1}{\mathrm{z}}& 3+\frac{1}{\mathrm{z}}\end{array}\right|\\ =\mathrm{xyz}\left(3+\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}+\frac{1}{\mathrm{z}}\right)\left|\begin{array}{ccc}1& 1& 1\\ 1+\frac{1}{\mathrm{y}}& 2+\frac{1}{\mathrm{y}}& \frac{1}{\mathrm{y}}\\ 1+\frac{1}{\mathrm{z}}& 1+\frac{1}{\mathrm{z}}& 3+\frac{1}{\mathrm{z}}\end{array}\right|\\ =\mathrm{xyz}\left(3+\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}+\frac{1}{\mathrm{z}}\right)\left|\begin{array}{ccc}1& 0& 0\\ 1+\frac{1}{\mathrm{y}}& 1& -1\\ 1+\frac{1}{\mathrm{z}}& 0& 2\end{array}\right|\\ =2\mathrm{xyz}\left(3+\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}+\frac{1}{\mathrm{z}}\right)\end{array}$

Hence, the given equation gives ${\mathrm{x}}^{-1}+{\mathrm{y}}^{-1}+{\mathrm{z}}^{-1}=-3$.