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 ?
 A: $\mathbf u = (\frac {\sqrt 2}{2} \mathbf i - \frac {\sqrt 2}{2} \mathbf j)\\
\mathbf v = (\frac {\sqrt 6}{6} \mathbf i + \frac {\sqrt 6}{6} \mathbf j -\frac {\sqrt 6}{3} \mathbf k)$
$(x,y,z) = (1,1,-1) + 3\mathbf u \cos t + 3\mathbf v \sin t\\ 
x = 1 + 3\frac {\sqrt 2}{2} \cos t + \frac{\sqrt {6}}{2} \sin t\\
y = 1 - 3\frac {\sqrt 2}{2} \cos t + \frac {\sqrt {6}}{2} \sin t\\
z = -1  - \sqrt {6} \sin t$
A: 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*}
A: The intersection of the sphere 
$$(x-1)^2+(y-1)^2+(z+1)^2=9$$
and the plane
$$x+y+z=1$$
can be parametrised by spherical coordinates :
$$x=1+3\sin (\phi)\cos (\theta) $$
$$y=1+3\sin (\phi)\sin (\theta) $$
$$z=-1-3\sin (\phi)(\cos (\theta)+\sin (\theta)) $$.
with $0\le \theta \le 2\pi $ and $0\le \phi \le \pi$.
