Answer
$${\text{The graphic description does not exist in real numbers}}{\text{.}}$$
Work Step by Step
$$\eqalign{
& {x^2} - 4x + {y^2} + 6y + {z^2} + 14 = 0 \cr
& {\text{Group terms}} \cr
& \left( {{x^2} - 4x} \right) + \left( {{y^2} + 6y} \right) + {z^2} = - 14 \cr
& {\text{Complete the square}} \cr
& \left( {{x^2} - 4x + 4} \right) + \left( {{y^2} + 6y + 9} \right) + {z^2} = - 14 + 4 + 9 \cr
& {\left( {x + 1} \right)^2} + {\left( {y - 1} \right)^2} + {z^2} = - 1 \cr
& {\text{It is not possible for all real numbers that }} \cr
& {\left( {x + 1} \right)^2} + {\left( {y - 1} \right)^2} + {z^2}{\text{ be a negative number, then}} \cr
& {\text{The graphic description does not exist in real numbers}}{\text{.}} \cr} $$
