We have:\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\)

\(=\dfrac{y+z+1+x+z+2+x+y-3}{x+y+z}=\dfrac{2x+2y+2z}{x+y+z}=2\)

\(\Rightarrow\left\{{}\begin{matrix}y+z+1=2x\left(\cdot\right)\\x+z+2=2y\left(\cdot\cdot\right)\\x+y-3=2z\left(\cdot\cdot\cdot\right)\\x+y+z=\dfrac{1}{2}\left(\cdot\cdot\cdot\cdot\right)\end{matrix}\right.\)

\(\left(\cdot\cdot\cdot\cdot\right)-\left(\cdot\right)\Rightarrow x-1=\dfrac{1}{2}-2x\Rightarrow3x=\dfrac{3}{2}\Rightarrow x=\dfrac{1}{2}\)

\(\left(\cdot\cdot\cdot\cdot\right)-\left(\cdot\cdot\right)\Rightarrow y-2=\dfrac{1}{2}-2y\Rightarrow3y=\dfrac{5}{2}\Rightarrow y=\dfrac{5}{6}\)

\(\left(\cdot\cdot\cdot\cdot\right)-\left(\cdot\cdot\cdot\right)\Rightarrow z+3=\dfrac{1}{2}-2z\Rightarrow3z=-\dfrac{5}{2}\Rightarrow z=-\dfrac{5}{6}\)

\(\Rightarrow P=2016x+y^{2017}+z^{2017}=2016.\dfrac{1}{2}+\left(\dfrac{5}{6}\right)^{2017}+\left(-\dfrac{5}{6}\right)^{2017}\)

\(=1008+\left(\dfrac{5}{6}\right)^{2017}-\left(\dfrac{5}{6}\right)^{2017}=1008\)

We have:y+z+1x=x+z+2y=x+y−3z=1x+y+z

=y+z+1+x+z+2+x+y−3x+y+z=2x+2y+2zx+y+z=2

⇒⎧⎩⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪y+z+1=2x(⋅)x+z+2=2y(⋅⋅)x+y−3=2z(⋅⋅⋅)x+y+z=12(⋅⋅⋅⋅)

(⋅⋅⋅⋅)−(⋅)⇒x−1=12−2x⇒3x=32⇒x=12

(⋅⋅⋅⋅)−(⋅⋅)⇒y−2=12−2y⇒3y=52⇒y=56

(⋅⋅⋅⋅)−(⋅⋅⋅)⇒z+3=12−2z⇒3z=−52⇒z=−56

⇒P=2016x+y2017+z2017=2016.12+(56)2017+(−56)2017

=1008+(56)2017−(56)2017=1008