Answer
$${\text{Center }}\left( {5, - 3,2} \right),{\text{ radius }}r = 2$$
Work Step by Step
$$\eqalign{
& {x^2} + {y^2} + {z^2} - 10x + 6y - 4z + 34 = 0 \cr
& {\text{Group terms}} \cr
& \left( {{x^2} - 10x} \right) + \left( {{y^2} + 6y} \right) + \left( {{z^2} - 4z} \right) = - 34 \cr
& {\text{Complete the square}} \cr
& \left( {{x^2} - 10x + 25} \right) + \left( {{y^2} + 6y + 9} \right) + \left( {{z^2} - 4z + 4} \right) = - 34 + 38 \cr
& {\text{Factor}} \cr
& {\left( {x - 5} \right)^2} + {\left( {y + 3} \right)^2} + {\left( {z - 2} \right)^2} = {\left( 2 \right)^2} \cr
& {\text{The standard equation of a sphere of radius }}r,{\text{ centered at}} \cr
& \left( {{x_0},{y_0},{z_0}} \right){\text{ is:}} \cr
& {\text{Then}} \cr
& {\text{Center }}\left( {5, - 3,2} \right),{\text{ radius }}r = 2 \cr} $$
