Answer
$(x - 1)^2+(y -2)^2+(z - 3)^2=14$
Work Step by Step
As we know the standard equation of sphere is $(x-a)^2+(y-b)^2+(z-c)^2=r^2$ ...(a)
where $(a,b,c)$ represents center and radius of the sphere is $r$
Plug $a=1,b=2,c=3, r=\sqrt{14}$ in equation (a).
This implies $(x - 1)^2+(y -2)^2+(z - 3)^2=(\sqrt{14})^2$
or, $(x - 1)^2+(y -2)^2+(z - 3)^2=14$
