Cho $(S):(x-1)^2+(y-1)^2+z^2=6$ và hai đường thẳng
$d_1:\left\{ \begin{array}{l} x=1+2t\\ y=3-2t\\z=1+2t \end{array} \right. ;     d_2:\left\{ \begin{array}{l} x=1-t'\\ y=2+2t'\\z=1-3t' \end{array} \right. $
Viết phương trình mặt phẳng tiếp xúc mặt cầu $(S)$ đồng thời song song với $d_1.d_2$
$d_1$ có vecto chỉ phương $\overrightarrow{a}=(2;-2;2) $
$d_2$ có vecto chỉ phương $\overrightarrow{b}=(-1;2;-3) $
Mặt cầu $(S)$ có tâm $I(1;1;0), R=\sqrt{6} $
Gọi $(\alpha )$ là mặt phẳng tiếp xúc mặt cầu $(S)$ và $(\alpha )$ song song với $d_1,d_2$
Vì $\left\{ \begin{array}{l} (\alpha )//d_1\\ (\alpha )//d_2 \end{array} \right. $
$\Rightarrow   (\alpha )$ có cặp vecto chỉ phương $\left\{ \begin{array}{l} \overrightarrow{a}=(2;-2;2) \\ \overrightarrow{b}=(-1;2;-3)  \end{array} \right. $
$\Rightarrow  (\alpha )$ có vecto pháp tuyến $[\overrightarrow{a}, \overrightarrow{b}  ]=(2;4;2)$

Chọn $\overrightarrow{n}=(1;2;1)\Rightarrow  (\alpha ):x+2y+z+D=0 $
$(\alpha )$ tiếp xức với $(S)$:
$\Rightarrow  d(I,(\alpha )) = R \Leftrightarrow \frac{{|{x_I} + 2{y_I} + {z_I} + D|}}{{\sqrt {1 + 4 + 1} }} = \sqrt 6  \Leftrightarrow \left[ \begin{array}{l}
3 + D = 6\\
3 + D = - 6
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
D = 3\\
D = - 9
\end{array} \right.$
Vậy phương trình mặt phẳng càn tìm: $x+2y+z+3=0$ hoặc $x+2y+z-9=0$

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