Answer
Center: $(-\dfrac{1}{4},-\dfrac{1}{4},-\dfrac{1}{4})$ and radius: $\dfrac{5 \sqrt 3}{4}$
Work Step by Step
We know that the standard equation of a sphere is written as:
$(x-x_0)^2+(y-y_0)^2+(z-z_0)^2=r^2\ \ \ $ (1)
Here, $(x_0,y_0,z_0)$ represents the center and $r$ is the radius of the sphere.
Since, we have $2x^2+2y^2+2z^2+x+y+z=9$
or, $2x^2+x+2y^2+y+2z^2+z=9$
or, $2(x^2+\dfrac{1}{2}x)+2(y^2+\dfrac{1}{2}y)+2(z^2+\dfrac{1}{2}z)=9$
or, $2(x^2+\dfrac{1}{2}x+\dfrac{1}{16})+2(y^2+\dfrac{1}{2}y+\dfrac{1}{16})+2(z^2+\dfrac{1}{2}z+\dfrac{1}{16})=9+\dfrac{2}{16}+\dfrac{2}{16}+\dfrac{2}{16}$
or, $(x +\dfrac{1}{4})^2+(y +\dfrac{1}{4})^2+(z +\dfrac{1}{4})^2=(\sqrt{\dfrac{75}{16}})^2$
or, $(x -(-\dfrac{1}{4}))^2+(y -(-\dfrac{1}{4}))^2+(z -(-\dfrac{1}{4}))^2=(\dfrac{5 \sqrt 3}{4})^2$
Compare the above equation with equation (1), we get
Center :$(-\dfrac{1}{4},-\dfrac{1}{4},-\dfrac{1}{4})$ and radius: $\dfrac{5 \sqrt 3}{4}$
