Trong không gian Oxyz, cho ba điểm $A\left(1;-2;3\right),B\left(4;2;3\right),C\left(3;4;3\right)$. Gọi $\left(S_1\right),\left(S_2\right),\left(S_3\right)$  là các mặt cầu có tâm A, B, C và bán kính lần lượt bằng 3, 2, 3. Hỏi có bao nhiêu mặt phẳng qua điểm $I\left(\frac{14}{5};\frac{2}{5};3\right)$ và tiếp xúc với cả 3 mặt cầu $\left(S_1\right),\left(S_2\right),\left(S_3\right)$.

Đáp án D.

Gọi $\overrightarrow{n}=\left(1;a;b\right)$ là 1 VTPT của (P), khi đó phương trình (P) là:

$1\left(x-\frac{14}{5}\right)+a\left(y-\frac{2}{5}\right)+b\left(z-3\right)=0\iff 5x+5\text{a}y+5b\text{z}-14-2a-15b=0$.

Theo bài ra ta có:

$\left\{\begin{array}{l}d\left(A;\left(P\right)\right)=3\\ d\left(B;\left(P\right)\right)=2\\ d\left(C;\left(P\right)\right)=3\end{array}\right.\iff \left\{\begin{array}{l}\frac{\left|5-10a+15b-14-2a-15b\right|}{\sqrt{25+25a^2+25b^2}}=3\\ \frac{\left|20+10a+15b-14-2a-15b\right|}{\sqrt{25+25a^2+25b^2}}=2\\ \frac{\left|15+20a+15b-14-2a-15b\right|}{\sqrt{25+25a^2+25b^2}}=3\end{array}\right.\iff \left\{\begin{array}{l}\frac{\left|-12a-9\right|}{5\sqrt{1+a^2+b^2}}=3\\ \frac{\left|8a+6\right|}{5\sqrt{1+a^2+b^2}}=2\\ \frac{\left|18a+1\right|}{5\sqrt{1+a^2+b^2}}=3\end{array}\right.$

$\iff \left\{\begin{array}{l}\frac{\left|4a+3\right|}{5\sqrt{1+a^2+b^2}}=1\\ \frac{\left|4a+3\right|}{5\sqrt{1+a^2+b^2}}=1\\ \frac{\left|18a+1\right|}{5\sqrt{1+a^2+b^2}}=3\end{array}\right.\iff \left\{\begin{array}{l}\left|4a+3\right|=5\sqrt{1+a^2+b^2}\\ \left|18a+1\right|=15\sqrt{1+a^2+b^2}\end{array}\right.\iff \left\{\begin{array}{l}\left|18a+1\right|=3\left|4a+3\right|\\ \left|4a+3\right|=5\sqrt{1+a^2+b^2}\end{array}\right.$

$\iff \left[\begin{array}{l}\left[\begin{array}{l}18a+1=12a+9\\ 18a+1=-12a-9\end{array}\right.\\ \left|4a+3\right|=5\sqrt{1+a^2+b^2}\end{array}\right.\iff \left\{\begin{array}{l}\left[\begin{array}{l}a=\frac{4}{3}\\ a=\frac{-1}{3}\end{array}\right.\\ \left|4a+3\right|=5\sqrt{1+a^2+b^2}\end{array}\right.\iff \left[\begin{array}{l}\left\{\begin{array}{l}a=\frac{4}{3}\\ \sqrt{\frac{25}{9}+b^2}=\frac{5}{3}\end{array}\right.\\ \left\{\begin{array}{l}a=\frac{-1}{3}\\ \sqrt{\frac{25}{9}+b^2}=\frac{1}{3}\left(vonghiem\right)\end{array}\right.\end{array}\right.\iff \left\{\begin{array}{l}a=\frac{4}{3}\\ b=0\end{array}\right.$ 

Vậy có 1 mặt phẳng thỏa mãn yêu cầu bài toán.