Answer
Center is $(0,-\dfrac{1}{3},\dfrac{1}{3})$ and radius is $\dfrac{\sqrt {29}}{3}$
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: $3x^2+3y^2+3z^2+2y-2z=9$
or, $2(x^2+\dfrac{1}{2}x)+2(y^2+\dfrac{1}{2}y)+2(z^2+\dfrac{1}{2}z)=9$
$ \implies x^2+(y+\dfrac{1}{3})^2+(z-\dfrac{1}{3})^2=\dfrac{29}{9}$
$ \implies (x-0)^2+(y -(-\dfrac{1}{3}))^2+(z-\dfrac{1}{3})^2=(\sqrt{\dfrac{29}{9}})^2$
and $(x-0)^2+(y -(-\dfrac{1}{3}))^2+(z-\dfrac{1}{3})^2=(\dfrac{\sqrt {29}}{3})^2$
Thus, we have
Center is $(0,-\dfrac{1}{3},\dfrac{1}{3})$ and radius is $\dfrac{\sqrt {29}}{3}$
