Cho đường thẳng \(d\): $\frac{{x - 1}}{2} = \frac{{y - 2}}{1} = \frac{{z - 3}}{2}$ và hai mặt phẳng $\left( {{P_1}} \right):\,\,x\, + \,2y\, + \,2z\, - \,2\, = \,0;\;\,$ $\left( {{P_2}} \right):\,\,2x\, + \,y\, + \,2z\, - \,1\, = \,0$. Mặt cầu có tâm \(I\) nằm trên \(d\) và tiếp xúc với 2 mặt phẳng \(\left( {{P_1}} \right),{\rm{ }}\left( {{P_2}} \right)\), có phương trình:
A.$\left( S \right):\,\,\,{\left( {x + 1} \right)^2} + {\left( {y + 2} \right)^2} + {\left( {z + 3} \right)^2} = 9.$
B.$\left( S \right):\,\,\,{\left( {x + 1} \right)^2} + {\left( {y + 2} \right)^2} + {\left( {z + 3} \right)^2} = 9$ hoặc $\left( S \right):\,\,\,{\left( {x + \frac{{19}}{{17}}} \right)^2} + {\left( {y + \frac{{16}}{{17}}} \right)^2} + {\left( {z + \frac{{15}}{{17}}} \right)^2} = \frac{9}{{289}}.$
C.$\left( S \right):\,\,\,{\left( {x - 1} \right)^2} + {\left( {y - 2} \right)^2} + {\left( {z - 3} \right)^2} = 9.$
D.$\left( S \right):\,\,\,{\left( {x - 1} \right)^2} + {\left( {y - 2} \right)^2} + {\left( {z - 3} \right)^2} = 9$ hoặc $\left( S \right):\,\,\,{\left( {x + \frac{{19}}{{17}}} \right)^2} + {\left( {y - \frac{{16}}{{17}}} \right)^2} + {\left( {z - \frac{{15}}{{17}}} \right)^2} = \frac{9}{{289}}.$
A.$\left( S \right):\,\,\,{\left( {x + 1} \right)^2} + {\left( {y + 2} \right)^2} + {\left( {z + 3} \right)^2} = 9.$
B.$\left( S \right):\,\,\,{\left( {x + 1} \right)^2} + {\left( {y + 2} \right)^2} + {\left( {z + 3} \right)^2} = 9$ hoặc $\left( S \right):\,\,\,{\left( {x + \frac{{19}}{{17}}} \right)^2} + {\left( {y + \frac{{16}}{{17}}} \right)^2} + {\left( {z + \frac{{15}}{{17}}} \right)^2} = \frac{9}{{289}}.$
C.$\left( S \right):\,\,\,{\left( {x - 1} \right)^2} + {\left( {y - 2} \right)^2} + {\left( {z - 3} \right)^2} = 9.$
D.$\left( S \right):\,\,\,{\left( {x - 1} \right)^2} + {\left( {y - 2} \right)^2} + {\left( {z - 3} \right)^2} = 9$ hoặc $\left( S \right):\,\,\,{\left( {x + \frac{{19}}{{17}}} \right)^2} + {\left( {y - \frac{{16}}{{17}}} \right)^2} + {\left( {z - \frac{{15}}{{17}}} \right)^2} = \frac{9}{{289}}.$
$I \in d\; \Rightarrow I\left( {2t + 1;\,t + 2;\,2t + 3} \right)$
Mặt cầu tiếp xúc với 2 mặt phẳng $ \Leftrightarrow d\left( {I;\left( {{P_1}} \right)} \right) = d\left( {I;\,\left( {{P_2}} \right)} \right)\,$
$ \Leftrightarrow \;\left| {8t + 9} \right| = \left| {9t + 9} \right| \Leftrightarrow \left[ \begin{array}{l}8t + 9 = 9t + 9\\8t - 9 = - 9t - 9\end{array} \right.\quad \Leftrightarrow \;\left[ \begin{array}{l}t = 0\\t = \frac{{ - 18}}{{17}}\end{array} \right.$
$t = 0 \Rightarrow \,I\left( {1;2;3} \right);\,\,R = 3 \Rightarrow \left( S \right):\,\,\,{\left( {x - 1} \right)^2} + {\left( {y - 2} \right)^2} + {\left( {z - 3} \right)^2} = 9.$
$t = - \frac{{18}}{{17}} \Rightarrow \,I\left( { - \frac{{19}}{{17}};\frac{{16}}{{17}};\frac{{15}}{{17}}} \right);\,\,R = \frac{3}{{17}} \Rightarrow \,\,\left( S \right):\,\,\,{\left( {x + \frac{{19}}{{17}}} \right)^2} + {\left( {y - \frac{{16}}{{17}}} \right)^2} + {\left( {z - \frac{{15}}{{17}}} \right)^2} = \frac{9}{{289}}.$
Lựa chọn đáp án D.
Mặt cầu tiếp xúc với 2 mặt phẳng $ \Leftrightarrow d\left( {I;\left( {{P_1}} \right)} \right) = d\left( {I;\,\left( {{P_2}} \right)} \right)\,$
$ \Leftrightarrow \;\left| {8t + 9} \right| = \left| {9t + 9} \right| \Leftrightarrow \left[ \begin{array}{l}8t + 9 = 9t + 9\\8t - 9 = - 9t - 9\end{array} \right.\quad \Leftrightarrow \;\left[ \begin{array}{l}t = 0\\t = \frac{{ - 18}}{{17}}\end{array} \right.$
$t = 0 \Rightarrow \,I\left( {1;2;3} \right);\,\,R = 3 \Rightarrow \left( S \right):\,\,\,{\left( {x - 1} \right)^2} + {\left( {y - 2} \right)^2} + {\left( {z - 3} \right)^2} = 9.$
$t = - \frac{{18}}{{17}} \Rightarrow \,I\left( { - \frac{{19}}{{17}};\frac{{16}}{{17}};\frac{{15}}{{17}}} \right);\,\,R = \frac{3}{{17}} \Rightarrow \,\,\left( S \right):\,\,\,{\left( {x + \frac{{19}}{{17}}} \right)^2} + {\left( {y - \frac{{16}}{{17}}} \right)^2} + {\left( {z - \frac{{15}}{{17}}} \right)^2} = \frac{9}{{289}}.$
Lựa chọn đáp án D.