Answer
$${\text{Center }}\left( {2,3,0} \right),{\text{ radius }}r = 3$$
Work Step by Step
$$\eqalign{
& {x^2} + {y^2} + {z^2} - 4x - 6y + 4 = 0 \cr
& {\text{Group terms}} \cr
& \left( {{x^2} - 4x} \right) + \left( {{y^2} - 6y} \right) + {z^2} = - 4 \cr
& {\text{Complete the square}} \cr
& \left( {{x^2} - 4x + 4} \right) + \left( {{y^2} - 6y + 9} \right) + {z^2} = - 4 + 4 + 9 \cr
& {\text{Factor}} \cr
& {\left( {x - 2} \right)^2} + {\left( {y - 3} \right)^2} + {z^2} = 9 \cr
& {\left( {x - 2} \right)^2} + {\left( {y - 3} \right)^2} + {\left( {z - 0} \right)^2} = {\left( 3 \right)^2} \cr
& {\text{The standard equation of a sphere of radius }}r,{\text{ centered at}} \cr
& \left( {{x_0},{y_0},{z_0}} \right){\text{ is:}} \cr
& {\text{Then}} \cr
& {\text{Center =}}\left( {2,3,0} \right),{\text{ radius }}r = 3 \cr} $$
