Answer
(a) $6$
(b) $(x-4)^2+(y-3)^2+(z+1)^2=36$
(c) $\langle 2, -4, 4 \rangle=2i-4j+4k$
Work Step by Step
We are given two points in the space as $P(4,3,-1)$ and $Q(6,-1,3)$
(a) The distance between the two points is given by $d=\sqrt {(6-4)^2+(-1-3)^2+(3+1)^2}=6$
(b) Sine $\vec{PQ}$ is the radius, we have $r=6$, with P as the center, the equation for the sphere is then
$(x-4)^2+(y-3)^2+(z+1)^2=36$
(c) Vector $\vec u=\vec{PQ}=\langle (6-4),(-1-3),(3+1) \rangle=\langle 2, -4, 4 \rangle=2i-4j+4k$
