In order to eliminate the first degree terms in the equation $2{\mathrm{x}}^2-2{\mathrm{y}}^2+{\mathrm{z}}^2-4\mathrm{x}+8\mathrm{y}+2\mathrm{z}-5=0$ the origin should be shifted to the point

$\mathrm{S}=2{\mathrm{x}}^2-2{\mathrm{y}}^2+{\mathrm{z}}^2-4\mathrm{x}+8\mathrm{y}+2\mathrm{z}=0 \mathrm{Solving} \frac{\partial S}{\partial x}=0, \frac{\partial S}{\partial y}=0,\frac{\partial S}{\partial z}=0$

or

$$\begin{gathered} \mathrm{The} \mathrm{equation} 2{\mathrm{x}}^2-2{\mathrm{y}}^2+{\mathrm{z}}^2-4\mathrm{x}+8\mathrm{y}+2\mathrm{z}-5=0 \mathrm{can} \mathrm{be} \mathrm{written} \mathrm{as} \\ 2(\mathrm{x}-1{)}^2-2(\mathrm{y}-2{)}^2+(\mathrm{z}+1{)}^2=0 \\ \mathrm{Let} \mathrm{x}-1=\mathrm{X},\mathrm{y}-2=\mathrm{Y},\mathrm{z}+1=\mathrm{Z} \\ \mathrm{then} \mathrm{the} \mathrm{given} \mathrm{equation} \mathrm{reduces} \mathrm{to} 2{\mathrm{X}}^2-2{\mathrm{Y}}^2+{\mathrm{Z}}^2=0 \\ \mathrm{which} \mathrm{do} \mathrm{not} \mathrm{contain} \mathrm{first} \mathrm{degree} \mathrm{terms} \\ \mathrm{Hence} \mathrm{coordinates} \mathrm{of} \mathrm{new} \mathrm{origin} =\left(1,2,-1\right) \end{gathered}$$