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 is given as:
$\quad (x-1)^{2}+(y-1)^{2}+(z-1)^{2}=1$
This equation reads as follows:
the square of the distance of a point on the sphere from its center point is 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:
$(x-1)^{2}+(y-1)^{2}+(z-1)^{2} \gt 1$
