Answer
Center is $(-\dfrac{1}{4},-\dfrac{1}{4},-\dfrac{1}{4})$ and radius is $\dfrac{5 \sqrt 3}{4}$
Work Step by Step
As we know the standard equation of sphere is $(x-a)^2+(y-b)^2+(z-c)^2=r^2$
where $(a,b,c)$ represents center and radius of the sphere is $r$
Given: $2x^2+2y^2+2z^2+x+y+z=9$
$\implies 2(x^2+\dfrac{1}{2}x)+2(y^2+\dfrac{1}{2}y)+2(z^2+\dfrac{1}{2}z)=9$
$ \implies 2(x^2+\dfrac{1}{2}x+\dfrac{1}{16})+2(y^2+\dfrac{1}{2}y+\dfrac{1}{16})+2(z^2+\dfrac{1}{2}z+\dfrac{1}{16})=9+\dfrac{2}{16}+\dfrac{2}{16}+\dfrac{2}{16}$
or, $(x +\dfrac{1}{4})^2+(y +\dfrac{1}{4})^2+(z +\dfrac{1}{4})^2=(\sqrt{\dfrac{75}{16}})^2$
and $(x -(-\dfrac{1}{4}))^2+(y -(-\dfrac{1}{4}))^2+(z -(-\dfrac{1}{4}))^2=(\dfrac{5 \sqrt 3}{4})^2$
Thus, we have Center is $(-\dfrac{1}{4},-\dfrac{1}{4},-\dfrac{1}{4})$ and radius is $\dfrac{5 \sqrt 3}{4}$
