Answer
$ a.\quad (x-1)^{2}+(y-1)^{2}+(z-1)^{2} \lt 1$
$ b.\quad (x-1)^{2}+(y-1)^{2}+(z-1)^{2} \gt 1$
Work Step by Step
The sphere:
$\quad (x-1)^{2}+(y-1)^{2}+(z-1)^{2}=1$
This equation states that the square of the distance of a point on the sphere from its centerpoint EQUALS 1
$ a.\quad$
If the point is inside the sphere, the distance is LESS than 1:
$(x-1)^{2}+(y-1)^{2}+(z-1)^{2} \lt 1$
$ b.\quad$
Outside the sphere, it is GREATER than 1
$(x-1)^{2}+(y-1)^{2}+(z-1)^{2} \gt 1$
