Answer
$(x-1)^2 +(y-2)^2+(z-2)^2= 4$
Work Step by Step
Let $(a,b,c)$ be the center of the sphere, $(x,y,z)$ be an arbitrary point of the sphere, and $r$ be the distance between $(a,b,c)$ and $(x,y,z)$. Then, using the Pythagorean Theorem, the standard form of the equation of the sphere can be expressed as:
$r=\sqrt {(x-a)^2+(y-b)^2+(z-c)^2}$
or, $(x-a)^2+(y-b)^2+(z-c)^2=r^2$
We are given that $ \ radius= r=1$ and center is at $(3,1,1)$
So, $(x-1)^2 +(y-2)^2+(z-2)^2=2^2$
Thus, the equation of the sphere is: $(x-1)^2 +(y-2)^2+(z-2)^2= 4$
