(S) có tâm $I\in d:\left\{ \begin{align}  & x=1-t \\  & y=-3+2t\left( t\in \mathbb{R} \right) \\  & 3+t \\ \end{align} \right.\Rightarrow I\left( 1-t;2t-3;t+3 \right)$

+) (S) Tiếp xúc với $\left( P \right)\Leftrightarrow R=d\left( I;\left( P \right) \right)\Leftrightarrow R=\frac{\left| 2\left( 1-t \right)+\left( 2t-3 \right)-2\left( t+3 \right)+9 \right|}{\sqrt{{{2}^{2}}+{{1}^{2}}+{{\left( -2 \right)}^{2}}}}=\frac{\left| 2-2t \right|}{3}$

+) (S) cắt (Q) theo một đường tròn (T) có chu vi $2\pi \Rightarrow \left( T \right)$ có bán kính r=1

+) $h=d\left( I;\left( Q \right) \right)=\frac{\left| \left( 1-t \right)-\left( 2t-3 \right)+\left( t+3 \right)+4 \right|}{\sqrt{{{1}^{2}}+{{\left( -1 \right)}^{2}}+{{1}^{2}}}}=\frac{\left| 11-2t \right|}{\sqrt{3}}$

Ta có ngay ${{R}^{2}}={{r}^{2}}+{{h}^{2}}\Rightarrow \frac{{{\left( 2-2t \right)}^{2}}}{9}=1+\frac{{{\left( 11-2t \right)}^{2}}}{3}\Leftrightarrow 4{{t}^{2}}-8t+4=9+3\left( 4{{t}^{2}}-44t+121 \right)$

$\Leftrightarrow 8{{t}^{2}}-124+368=0\Leftrightarrow \left[ \begin{align}  & t=4 \\  & t=\frac{23}{2} \\ \end{align} \right.$

Với $t=4\Rightarrow I\left( -3;5;7 \right)$và $R=2\Rightarrow \left( S \right):{{\left( x+3 \right)}^{2}}+{{\left( y-5 \right)}^{2}}+{{\left( z-7 \right)}^{2}}=4$