Intersection of a sphere and a plane knowing equations

We know the normal n of the plane and the radius of the circle. Call two vectors that are orthogonal to n and orthogonal to each other u and v. Then the circle is given by

 gamma(t) = I + n + r * cos(t) * u + r* sin(t) * v,

where n fulfills the length constraint n^2+r^2=R^2, where R is the radius of the sphere and r the radius of the circle. Your vector n satisfies this constraint already, so we do not have to change its normalization.

Now we need to figure out what the visible and invisible stretches are. Any point on the sphere has either a positive or negative projection on the normal vector on the screen

n_screen =({sin(\tdplotmaintheta)*sin(\tdplotmainphi)},{-sin(\tdplotmaintheta)*cos(\tdplotmainphi)},{cos(\tdplotmaintheta)});

So we need to find the zeros of the projection gamma(t).n_screen. This can be accomplished by letting TikZ find the intersections. Of course we do not really draw the paths here, and use overlay for them not to screw up the bounding box. Notice that the current version assumes that there are two zeros, so if you drastically change the view angles, this version won't work any more.

\documentclass[12pt,border=2mm,tikz]{standalone}
\usepackage{tikz-3dplot}
\usetikzlibrary{arrows,calc,backgrounds,intersections}
\begin{document}
\tdplotsetmaincoords{60}{110}
\begin{tikzpicture}[tdplot_main_coords,
declare function={dicri(\t,\th,\ph,\R)=
sin(\th)*sin(\ph)*(2+\R*cos(\t)*1/sqrt(2)+\R*sin(\t)*1/sqrt(18))-
sin(\th)*cos(\ph)*(-2+\R*cos(\t)*1/sqrt(2)-\R*sin(\t)*1/sqrt(18))+
cos(\th)*(1-4*\R*sin(\t)*1/sqrt(18));}]
\path
      coordinate (T) at (3,-3,3)
      coordinate (I) at (1,-1,2)
      coordinate (n) at (2,-2,1)
      coordinate (u) at ({1/sqrt(2)},{1/sqrt(2)},0)
      coordinate (v) at ({1/sqrt(18)},{-1/sqrt(18)},{-4/sqrt(18)});
      % the coordinatesn, u and v are not really used here

\foreach \v/\position in {T/above,I/below} {
    \draw[fill=black] (\v) circle (0.7pt) node [\position=0.2mm] {$\v$};
}
% \draw[red,thick,-latex] (0,0,0) --
% ({sin(\tdplotmaintheta)*sin(\tdplotmainphi)},
% {-sin(\tdplotmaintheta)*cos(\tdplotmainphi)},{cos(\tdplotmaintheta)});
% normal to screen
\begin{scope}[tdplot_screen_coords, on background layer]
  \pgfmathsetmacro{\R}{5}%
  \fill[ball color=purple, opacity=1.0] (I) circle (\R);
  % determine the zeros of dicri
  \path[overlay,name path=dicri] plot[variable=\x,domain=0:360,samples=73]
  ({\x*1pt},{dicri(\x,\tdplotmaintheta,\tdplotmainphi,4)});
  \path[overlay,name path=zero] (0,0) -- (360pt,0);
  \path[name intersections={of=dicri and zero,total=\t}]
  let \p1=(intersection-1),\p2=(intersection-2) in
  \pgfextra{\xdef\xmin{\x1}\xdef\xmax{\x2}};
\end{scope}
\pgfmathsetmacro{\R}{4}
\draw[dashed] plot[variable=\t,domain=\xmin:\xmax,samples=73,smooth] 
 ({1+2+\R*cos(\t)*1/sqrt(2)+\R*sin(\t)*1/sqrt(18)},
 {-1-2+\R*cos(\t)*1/sqrt(2)-\R*sin(\t)*1/sqrt(18)},
 {2+1-4*\R*sin(\t)*1/sqrt(18)});
\draw[thick] plot[variable=\t,domain=\xmax:\xmin+360,samples=73,smooth] 
 ({1+2+\R*cos(\t)*1/sqrt(2)+\R*sin(\t)*1/sqrt(18)},
 {-1-2+\R*cos(\t)*1/sqrt(2)-\R*sin(\t)*1/sqrt(18)},
 {2+1-4*\R*sin(\t)*1/sqrt(18)});
\end{tikzpicture}
\end{document}

And here is a plane, using your nicer vectors u and v from the chat.

\documentclass[12pt,border=2mm,tikz]{standalone} 
\usepackage{tikz-3dplot} 
\usetikzlibrary{arrows,calc,backgrounds,intersections} 
\makeatletter % https://tex.stackexchange.com/a/38995/121799
\tikzset{
  use path/.code={\pgfsyssoftpath@setcurrentpath{#1}}
}
\makeatother
\begin{document} 
\tdplotsetmaincoords{60}{110} 
\begin{tikzpicture}[tdplot_main_coords, 
  declare function={dicri(\t,\th,\ph,\R)=% 
  sin(\th)*sin(\ph)*(2+\R*cos(\t)/3+2*\R*sin(\t)/3)-%
  sin(\th)*cos(\ph)*(-2 +2*\R*cos(\t)/3 + \R*sin(\t)/3)+%
  cos(\th)*(1+2*\R*cos(\t)/3-2*\R*sin(\t)/3);}] 
  \pgfmathsetmacro{\R}{5}% 
  \path  coordinate (T) at (3,-3,3) 
   coordinate (I) at (1,-1,2) 
   coordinate (n) at (2,-2,1) 
   coordinate (u) at (1, 2, 2) 
   coordinate (v) at (2, 1, -2); 
  % the coordinatesn, u and v are not really used here 
   \path[tdplot_screen_coords,shift={(I)},use as bounding box] (-1.2*\R,-1.2*\R)rectangle (1.2*\R,1.2*\R);   

  \foreach \v/\position in {T/above,I/below} { 
   \draw[fill=black] (\v) circle (0.7pt) node [\position=0.2mm] {$\v$}; 
  } 
  % \draw[red,thick,-latex] (0,0,0) -- 
  % ({sin(\tdplotmaintheta)*sin(\tdplotmainphi)}, 
  % {-sin(\tdplotmaintheta)*cos(\tdplotmainphi)},{cos(\tdplotmaintheta)}); 
  % normal to screen 
  \begin{scope}[tdplot_screen_coords, on background layer] 
   \fill[ball color=green, opacity=0.8] (I) circle (\R); 
   % determine the zeros of dicri 
   \path[overlay,name path=dicri] plot[variable=\x,domain=0:360,samples=73] 
   ({\x*1pt},{dicri(\x,\tdplotmaintheta,\tdplotmainphi,4)}); 
   \path[overlay,name path=zero] (0,0) -- (360pt,0); 
   \path[name intersections={of=dicri and zero,total=\t}] 
   let \p1=(intersection-1),\p2=(intersection-2) in 
   \pgfextra{\xdef\tmin{\x1}\xdef\tmax{\x2}}; 
  \end{scope} 
  \pgfmathsetmacro{\SmallR}{4} 
  \draw[dashed] plot[variable=\t,domain=\tmin:\tmax,samples=50,smooth] 
   ({1+2+\SmallR*cos(\t)/3+2*\SmallR*sin(\t)/3}, 
   {-1-2 +2*\SmallR*cos(\t)/3+ \SmallR*sin(\t)/3}, 
   {2+1+2*\SmallR*cos(\t)/3 - 2*\SmallR*sin(\t)/3 }); 
  \draw[thick,save path=\pathA] plot[variable=\t,domain=\tmax:\tmin+360,samples=50,smooth] 
   ({1+2+\SmallR*cos(\t)/3+2*\SmallR*sin(\t)/3}, 
   {-1-2 +2*\SmallR*cos(\t)/3+ \SmallR*sin(\t)/3}, 
   {2+1+2*\SmallR*cos(\t)/3 - 2*\SmallR*sin(\t)/3 }); 
  \path ({1+2+\SmallR*cos(\tmin)/3+2*\SmallR*sin(\tmin)/3}, 
   {-1-2 +2*\SmallR*cos(\tmin)/3+ \SmallR*sin(\tmin)/3}, 
   {2+1+2*\SmallR*cos(\tmin)/3 - 2*\SmallR*sin(\tmin)/3 }) coordinate (pmin)
   ({1+2+\SmallR*cos(\tmax)/3+2*\SmallR*sin(\tmax)/3}, 
   {-1-2 +2*\SmallR*cos(\tmax)/3+ \SmallR*sin(\tmax)/3}, 
   {2+1+2*\SmallR*cos(\tmax)/3 - 2*\SmallR*sin(\tmax)/3 }) coordinate (pmax);
  \begin{scope}[tdplot_screen_coords]
   \clip[shift={(I)}] (-1.2*\R,-1.2*\R)rectangle (1.2*\R,1.2*\R);   
   \path[fill=gray,fill opacity=0.4,even odd rule] let \p1=($(pmin)-(I)$),\p2=($(pmax)-(I)$),
   \p3=($(pmax)-(pmin)$),\n1={atan2(\y1,\x1)},\n2={atan2(\y2,\x2)},
   \n3={atan2(\y3,\x3)}
    in [use path=\pathA]  (pmin) arc(\n1:\n2-360:\R) 
    (0,-6) -- ++(\n3:{12cm/sin(\n3)}) -- ++(\n3+90:{12cm/sin(\n3)})
    -- ++(\n3+180:{12cm/sin(\n3)}) -- cycle;
  \end{scope}
\end{tikzpicture} 
\end{document}

Many thanks to Schrödinger's cat about 3dtools. In this answer, I use Mathematica to find coordinates of three points A, B, C and draw circle (ABC). I use tikz-3dplot-circleofsphere at here to draw style of line of the circle (ABC).

\documentclass[12pt,tikz,border=2 mm]{standalone}
\usepackage{tikz-3dplot-circleofsphere}
\usetikzlibrary{3dtools}
\begin{document}
    \tdplotsetmaincoords{60}{110}
\begin{tikzpicture}[scale=1,tdplot_main_coords,declare function={R=5;
}]
    \path   (3,-3,3) coordinate (T)
    (1,-1,2) coordinate (I) 
    (1, {1/5 *(-23 - 2* sqrt(11))}, {1/5 *(19 - 4 *sqrt(11))}) coordinate (A)
    (1, {1/5 *(-23 + 2* sqrt(11))}, {1/5 *(19 + 4 *sqrt(11))}) coordinate (B)
    ({1/4* (13 + sqrt(119))}, {1/4 *(-13 + sqrt(119))}, 2) coordinate (C);
\begin{scope}[tdplot_screen_coords] 
\fill[ball color=green, opacity=0.8] (I) circle (R);
    \end{scope}
\begin{scope}[shift={(I)}]
\path[overlay] [3d coordinate={(A-B)=(A)-(B)},
3d coordinate={(A-C)=(A)-(C)},
3d coordinate={(myn)=(A-B)x(A-C)},
3d coordinate={(A-T)=(A)-(T)}];
\pgfmathsetmacro{\myaxisangles}{axisangles("(myn)")}
\pgfmathsetmacro{\myalpha}{{\myaxisangles}[0]}
\pgfmathsetmacro{\mybeta}{{\myaxisangles}[1]}
\pgfmathsetmacro{\mygamma}{-acos(sqrt(TD("(A-T)o(A-T)"))/R)}
\tdplotCsDrawCircle[tdplotCsFront/.style={thick}]{R}{\myalpha}{\mybeta}{\mygamma} 
\end{scope} 

\foreach \p in {I,T,B}
\draw[fill=black] (\p) circle (1.5 pt);
\foreach \p/\g in {I/0,T/-90,B/30}
\path (\p)+(\g:3mm) node{$\p$};
\draw[dashed] (I) -- (T) -- (B) -- cycle;
    \end{tikzpicture}
\end{document} 

I add this because you asked me to use the 3d circle through 3 points pic here. It can be used after we have three points. These points can be constructed by adding vectors of length 4, the radius, to T which lie in the plane of the circle. These vectors are taken from the other answer, and rescaled.

\documentclass[12pt,tikz,border=2 mm]{standalone}
\usepackage{tikz-3dplot}
\usetikzlibrary{3dtools,backgrounds}
\begin{document}
\tdplotsetmaincoords{60}{110}
\begin{tikzpicture}[tdplot_main_coords]
\pgfmathsetmacro{\mysq}{4/3}
\path   (3,-3,3) coordinate (T)
      (1,-1,2) coordinate (I) 
      (1, 2, 2) coordinate (u)  
      (2, 1, -2) coordinate (v)
      [3d coordinate={(A)=(T)+\mysq*(u)}]
      [3d coordinate={(B)=(T)+\mysq*(v)}]
      [3d coordinate={(C)=(T)-\mysq*(u)}];

\foreach \v/\position in {T/above,I/below} {
    \draw[fill=black] (\v) circle (0.7pt) node [\position=0.2mm] {$\v$};
}

\path[dashed] pic{3d circle through 3 points={A={(A)},B={(B)},C={(C)}}};

\begin{scope}[tdplot_screen_coords, on background layer]
  \pgfmathsetmacro{\R}{5}%
  \fill[ball color=purple, opacity=1.0] (I) circle (\R); 
\end{scope}
\end{tikzpicture}
\end{document}