Câu hỏi 48

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

\(1\left( {x - \dfrac{{14}}{5}} \right) + a\left( {y - \dfrac{2}{5}} \right) + b\left( {z - 3} \right) = 0 \Leftrightarrow 5x + 5ay + 5bz - 14 - 2a - 15b = 0\).

Theo bài ra ta có:

\(\begin{array}{l}\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. \Leftrightarrow \left\{ \begin{array}{l}\dfrac{{\left| {5 - 10a + 15b - 14 - 2a - 15b} \right|}}{{\sqrt {25 + 25{a^2} + 25{b^2}} }} = 3\\\dfrac{{\left| {20 + 10a + 15b - 14 - 2a - 15b} \right|}}{{\sqrt {25 + 25{a^2} + 25{b^2}} }} = 2\\\dfrac{{\left| {15 + 20a + 15b - 14 - 2a - 15b} \right|}}{{\sqrt {25 + 25{a^2} + 25{b^2}} }} = 3\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\dfrac{{\left| { - 12a - 9} \right|}}{{5\sqrt {1 + {a^2} + {b^2}} }} = 3\\\dfrac{{\left| {8a + 6} \right|}}{{5\sqrt {1 + {a^2} + {b^2}} }} = 2\\\dfrac{{\left| {18a + 1} \right|}}{{5\sqrt {1 + {a^2} + {b^2}} }} = 3\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}\dfrac{{\left| {4a + 3} \right|}}{{5\sqrt {1 + {a^2} + {b^2}} }} = 1\\\dfrac{{\left| {4a + 3} \right|}}{{5\sqrt {1 + {a^2} + {b^2}} }} = 1\\\dfrac{{\left| {18a + 1} \right|}}{{5\sqrt {1 + {a^2} + {b^2}} }} = 3\end{array} \right. \Leftrightarrow \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. \Leftrightarrow \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.\\ \Leftrightarrow \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. \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}a = \dfrac{4}{3}\\a = \dfrac{{ - 1}}{3}\end{array} \right.\\\left| {4a + 3} \right| = 5\sqrt {1 + {a^2} + {b^2}} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}a = \dfrac{4}{3}\\\sqrt {\dfrac{{25}}{9} + {b^2}}  = \dfrac{5}{3}\end{array} \right.\\\left\{ \begin{array}{l}a =  - \dfrac{1}{3}\\\sqrt {\dfrac{{25}}{9} + {b^2}}  = \dfrac{1}{3}\,\,\left( {vo\,\,nghiem} \right)\end{array} \right.\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = \dfrac{4}{3}\\b = 0\end{array} \right.\end{array}\)

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

Chọn D.