$4x^2+2y^2+2z^2-4xy-4xz-2yz-6y-10z+34=0$
$⇒4x^2-4xy+y^2-2z(2x+y)+z^2+y^2-6y+9+z^2-10z+25=0$
$⇒(2x-y)^2-2z(2x+y)+z^2+(y-3)^2+(z-5)^2=0$
$⇒(2x-y-z)^2+(y-3)^2+(z-5)^2=0$
Vì: $(2x-y+z)^2+(y-3)^2+(z-5)^2≥0(∀x;y;z)$
$⇒$ $\begin{cases} 2x-y-z=0\\y-3=0\\z-5=0 \end{cases}$
$⇒$ $\begin{cases} 2x-3-5=0\\y=3\\z=5 \end{cases}$
$⇒$ $\begin{cases} 2x=8\\y=3\\z=5 \end{cases}$
$⇒$ $\begin{cases} x=4\\y=3\\z=5 \end{cases}$
Thay vào $S$:
$S=(4-4)^{2020}+(3-4)^{2020}+(5-4)^{2020}$
$=0+1+1$
$=2$
Vậy $K=2$