If a curve $x=x(s)$ lies on the sphere whose radius is $a$ and center is at $c$ and whose equation is $(y-c).(y-c)=a^2$, then show that $$ \begin{align} &(i)\quad c=x+\frac1\kappa n-\frac{\kappa '}{\kappa^2\tau }b\\ &(ii)\quad \frac{d}{ds}\left(\frac{\kappa '}{\kappa^2 \tau}\right)-\frac{\tau}{\kappa}=0\\ &(iii)\quad \left(\frac{1}{\kappa}\right)^2+\left(\frac{\kappa '}{\kappa^2 \tau}\right)^2=a^2 \end{align} $$ Where $t,n,b,\kappa$ and $\tau$ are tangent, normal, binormal, curvature and torsion respectively.
From serret-Frenet formulae, $$ \begin{align} t' &=\kappa n\\ n' &= -\kappa t + \tau b\\ b' &= \tau n \end{align} $$
Since $x=x(s)$ lies on the sphere, I start with differentiating the equation,
$$ \begin{align} (x-c).(x-c)&=a^2\\ 2x'.(x-c)&=0\\ t.(x-c)&=0\\ x't+(x-c).t'&=0\quad\text{differentiating again w.r.t. }s\\ t.t+(x-c).\kappa n &=0\\ (x-c).n &= - \frac{1}{\kappa} \end{align} $$
Now I couldn't understand how to go further. Any help will be appreciated.
Thanks in advance.
You have already obtained
[1] $(x-c).t = 0$
[2] $(x-c).n = -1/κ$
Differenciating the equation [2],
[3] $(x-c).b = κ'/κ^2τ$
As in my comment, these results and the property of $t,n,b$ being orthonormal system imply
[4] $x-c = (-1/κ)n + (κ'/κ^2τ)b$,
which is equivalent to $(i)$.
Further differenciating the equation [3],
[5] $(x-c).(-τn) = d/ds(κ'/κ^2τ)$,
which imply $(ii)$ using [2].
(Here your equation $b'=τn$ is corrected to $b'=-τn$.)
Finally, evaluating $(x-c).(x-c)=a^2$ using [4] will show $(iii)$.