Answer
The sphere is centered at $(0,1,2)$ and has radius $\frac{5}{\sqrt{3}}$.
Work Step by Step
To show the equation $3x^2+3y^2+3z^2=10+6y+12z$ is an equation of a sphere, we will put it in the form $(x-a)^2+(y-b)^2+(z-c)^2=r^2$, where $a$, $b$, $c$, and $r$ are real numbers.
First, we "move" both of the variable terms from the right hand side of the equation to the left hand side, then we divide both sides by 3. We get
$$x^2+y^2-2y+z^2-4z=\frac{10}{3}.$$
Next we want to complete the squares of $y^2-2y$ and $z^2-4z$, so we add 1 and 4 to both sides, then rearrange the left hand side:
$$x^2+y^2-2y+z^2-4z+1+4=\frac{10}{3}+1+4
\\x^2+(y^2-2y+1)+(z^2-4z+4)=\frac{25}{3}
\\x^2+(y-1)^2+(z-2)^2=\frac{25}{3}.$$
Hence, this is a sphere centered at $(0,1,2)$ and with radius $\sqrt{\frac{25}{3}}=\frac{5}{\sqrt{3}}.$