Cho đường thẳng d x-1/2=y-2/1

* Hướng dẫn giải

Đáp án D

• $I\in d\Rightarrow I\left(2t+1;t+2;2t+3\right)$
• Mặt cầu tiếp xúc với 2 mặt phẳng $\iff d\left(I;\left(P_1\right)\right)=d\left(I_2;\left(P_2\right)\right)$$\iff \left|8t+9\right|=\left|9t+9\right|\iff \left[\begin{array}{l}8t+9=9t+9\\ 8t-9=-9t-9\end{array}\right.\iff \left[\begin{array}{l}t=0\\ t=\frac{-18}{17}\end{array}\right.$
• $t=0\Rightarrow I\left(1;2;3\right);R=3\Rightarrow \left(S\right):{\left(x-1\right)}^2+{\left(y-2\right)}^2+{\left(z-3\right)}^2=9$.

• $$\begin{gathered} t=-\frac{18}{17}\Rightarrow I\left(-\frac{19}{17};\frac{16}{17};\frac{15}{17}\right);R=\frac{3}{17} \\ \Rightarrow \left(S\right):{\left(x+\frac{19}{17}\right)}^2+{\left(y-\frac{16}{17}\right)}^2+{\left(z-\frac{15}{17}\right)}^2=\frac{9}{289} \end{gathered}$$.