The number of ordered triplet(s)  $(x,y,z)$ where $x,y,z\in R^{+}$  satisfying the following equations is/are

$\begin{array}{l}2x-2y+\frac{1}{z}=\frac{1}{2023}\\ 2y-2z+\frac{1}{x}=\frac{1}{2023}\\ 2z-2x+\frac{1}{y}=\frac{1}{2023}\end{array}$

Now, $2xz-2yz+1=\frac{z}{2023}$  -----------------------------------(1)
and $2yx-2zx+1=\frac{x}{2023}$  ------------------------------------(2)
and $2zy-2xy+1=\frac{y}{2023}$ -------------------------------------(3)
Adding eq.s (1),(2) and (3), we get  $3=\frac{z+x+y}{2023}$
i.e., $\text{x}+\text{y}+\text{z}=\text{3}$  (2023) -------------------------------------------(4)          
Similarly by adding given expressions, 
we get  $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{3}{2023}$-------------------(5)
Now by Cauchy-Schwartz inequality,  $(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\ge {\left(3\right)}^2$
i.e.,  $3\left(2023\right)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\ge 9\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge \frac{9}{3.2023}\ge \frac{3}{2023}$
But, $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{3}{2023}$ (from eq.(5))
Hence, equality should hold  $\Rightarrow x=y=z$
As,   $x+y+z=3\left(2023\right)\Rightarrow x=2023;y=2023;z=2023.$