Parametrise the circle centered at $ \ (1,1,-1) \ $ with radius equal to $ 3 $ Parametrise the circle centered at $ \ (1,1,-1) \ $ with radius equal to $ 3  $ in the plane $ x+y+z=1 $ with positive orientation . $$ $$ I have thought the parametriation: 
\begin{align} x(t)=1+ 3 \cos (t) \hat j +3 \sin (t)  \hat k \\ y(t)=1+3 \cos (t) \hat i+3 \sin (t) \hat k \\ z(t)=-1+3 \cos (t) \hat i +3 \sin (t) \hat j  , \ \ 0 \leq t \leq 2 \pi \end{align}  But I am not sure . Any help is there ?

The plane can be parameterised as follows
\begin{eqnarray*}
x=1+t \\
y=1+s \\
z=-1-t-s.
\end{eqnarray*}
Now substitute this into the equation for the sphere $(x-1)^2+(y-1)^2+(z+1)^2=9$. We have
\begin{eqnarray*}
t^2+s^2+(t+s)^2=9 \\
\end{eqnarray*}
Rearrange this to
\begin{eqnarray*}
(2t+s)^2+3s^2=18
\end{eqnarray*}
This can be paramterised 
\begin{eqnarray*}
s=\sqrt{6} \sin(\theta) \\
2t+s= \sqrt{18} \cos(\theta)
\end{eqnarray*}
Now substitute back into the parameterisation of the plane
\begin{eqnarray*}
x=1+ \frac{\sqrt{18} \cos(\theta) - \sqrt{6} \sin(\theta)}{2} \\
y=1+\sqrt{6} \sin(\theta) \\
z=-1-\frac{\sqrt{18} \cos(\theta)+\sqrt{6} \sin(\theta)}{2}
\end{eqnarray*}
