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$ \sqrt{x^2-2x+6}\cdot\log_{3}(6-y) = x $

$ \sqrt{y^2-2y+6}\cdot\log_{3}(6-z) = y $

$ \sqrt{z^2-2z+6}\cdot\log_{3}(6-x) = z $

My approach :: Here $6-x,6-y,6-z>0$. So $x,y,z<6$

Now $\displaystyle \log_{3}(6-y) = \frac{x}{\sqrt{x^2-2x+6}}=\frac{x}{\sqrt{(x-1)^2+5}}$

and $\displaystyle \log_{3}(6-z) = \frac{y}{\sqrt{y^2-2y+6}}=\frac{y}{\sqrt{(y-1)^2+5}}$

and $\displaystyle \log_{3}(6-x) = \frac{z}{\sqrt{z^2-2z+6}}=\frac{z}{\sqrt{(z-1)^2+5}}$

How can I calculate after that.

Help please

Thanks