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Given the determinant: $$\Delta = \begin{vmatrix} 1+x & 1 & 1 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z \end{vmatrix}$$ Factor out $x$, $y$, and $z$ from the first, second, and third rows respectively: $$\Delta = xyz \begin{vmatrix} \frac{1}{x}+1 & \frac{1}{x} & \frac{1}{x} \\ \frac{1}{y} & \frac{1}{y}+1 & \frac{1}{y} \\ \frac{1}{z} & \frac{1}{z} & \frac{1}{z}+1 \end{vmatrix}$$
Perform the operation $C_1 \rightarrow C_1 + C_2 + C_3$: $$\Delta = xyz \begin{vmatrix} 1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} & \frac{1}{x} & \frac{1}{x} \\ 1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} & \frac{1}{y}+1 & \frac{1}{y} \\ 1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} & \frac{1}{z} & \frac{1}{z}+1 \end{vmatrix}$$ Let $k = 1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}$. Factor out $k$ from $C_1$: $$\Delta = xyz \cdot k \begin{vmatrix} 1 & \frac{1}{x} & \frac{1}{x} \\ 1 & \frac{1}{y}+1 & \frac{1}{y} \\ 1 & \frac{1}{z} & \frac{1}{z}+1 \end{vmatrix}$$
Using row operations $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$: $$\Delta = xyz(1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}) \begin{vmatrix} 1 & \frac{1}{x} & \frac{1}{x} \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} = xyz(1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z})$$ Expanding, we get $\Delta = xyz + yz + xz + xy$.
If $\Delta = 0$, then $xyz(1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}) = 0$. Since $x, y, z \neq 0$, then $xyz \neq 0$. Therefore, $1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$, which implies $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = -1$.
Final Answer: \Delta = xyz + xy + yz + zx; \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = -1
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