Đáp án: −4.

Ta có \(\overrightarrow {CD} = \left( {4;2; - 4} \right)\).

Gọi phương trình mặt phẳng cần tìm là \(\left( P \right):x + ay + bz + c = 0.\) Suy ra \(\overrightarrow {{n_P}} = \left( {1;a;b} \right)\).

Vì \(\left( P \right){\rm{//}}CD\) nên \(\overrightarrow {{n_P}} \cdot \overrightarrow {CD} = 0\)\( \Leftrightarrow 4 + 2a - 4b = 0\)\( \Leftrightarrow 2 + a - 2b = 0\) (1).

Vì \(\left( P \right)\) tiếp xúc với hai mặt cầu \(\left( {{S_1}} \right),\left( {{S_2}} \right)\) nên \(\left\{ \begin{array}{l}d\left( {A,\left( P \right)} \right) = 3\\d\left( {B,\left( P \right)} \right) = \frac{3}{2}\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}\left| {1 + 2a - 3b + c} \right| = 3\sqrt {{1^2} + {a^2} + {b^2}} \\\left| {\frac{3}{2} + \frac{3}{2}a - \frac{1}{2}b + c} \right| = \frac{3}{2}\sqrt {{1^2} + {a^2} + {b^2}} \end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}\left| {1 + 2a - 3b + c} \right| = 3\sqrt {{1^2} + {a^2} + {b^2}} \\\left| {3 + 3a - b + 2c} \right| = 3\sqrt {{1^2} + {a^2} + {b^2}} \end{array} \right.\) (2).

Từ (1) và (2), ta có hệ phương trình \[ \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{2 + a - 2b = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\\begin{array}{l}\left| {1 + 2a - 3b + c} \right| = \left| {3 + 3a - b + 2c} \right|\\\left| {3 + 3a - b + 2c} \right| = 3\sqrt {{1^2} + {a^2} + {b^2}} \end{array}\end{array}} \right.\]

\[ \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{a = 2b - 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\\begin{array}{l}\left| {1 + 4b - 4 - 3b + c} \right| = \left| {3 + 6b - 6 - b + 2c} \right|\\\left| {3 + 6b - 6 - b + 2c} \right| = 3\sqrt {{1^2} + {{\left( {2b - 2} \right)}^2} + {b^2}} \end{array}\end{array}} \right.\]\[ \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{a = 2b - 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\\begin{array}{l}\left| {b + c - 3} \right| = \left| {5b + 2c - 3} \right|\\\left| {3 + 6b - 6 - b + 2c} \right| = 3\sqrt {{1^2} + {{\left( {2b - 2} \right)}^2} + {b^2}} \end{array}\end{array}} \right.\].

TH1: \[\left\{ {\begin{array}{*{20}{c}}{a = 2b - 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\\begin{array}{l}b + c - 3 = 5b + 2c - 3\\\left| {3 + 6b - 6 - b + 2c} \right| = 3\sqrt {{1^2} + {{\left( {2b - 2} \right)}^2} + {b^2}} \end{array}\end{array}} \right.\]\[ \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{a = 2b - 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\\begin{array}{l}c = - 4b\\\left| {3 + 6b - 6 - b - 8b} \right| = 3\sqrt {{1^2} + {{\left( {2b - 2} \right)}^2} + {b^2}} \end{array}\end{array}} \right.\]

\[ \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{a = 2b - 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\\begin{array}{l}c = - 4b\\\left| {b + 1} \right| = \sqrt {5{b^2} - 8b + 5} \end{array}\end{array}} \right.\]\[ \Leftrightarrow \left\{ \begin{array}{l}a = 2b - 2\\c = - 4b\\4{b^2} - 10b + 4 = 0\end{array} \right.\]\[ \Leftrightarrow \left\{ \begin{array}{l}a = 2b - 2\\c = - 4b\\b = 2\end{array} \right.\] hoặc \[\left\{ {\begin{array}{*{20}{c}}{a = 2b - 2}\\\begin{array}{l}c = - 4b\,\,\,\,\,\\b = \frac{1}{2}\end{array}\end{array}} \right.\]

\[ \Leftrightarrow \left\{ \begin{array}{l}a = 2\\c = - 8\\b = 2\end{array} \right.\] hoặc \[\left\{ {\begin{array}{*{20}{c}}{a = - 1}\\\begin{array}{l}c = - 2\\b = \frac{1}{2}\end{array}\end{array}} \right.\].

Với \[\left\{ \begin{array}{l}a = 2\\c = - 8\\b = 2\end{array} \right.\] thì \(\left( P \right):x + 2y + 2z - 8 = 0\) (thỏa mãn). Suy ra \(a + b + c = - 4.\)

Với \[\left\{ {\begin{array}{*{20}{c}}{a = - 1}\\\begin{array}{l}c = - 2\\b = \frac{1}{2}\end{array}\end{array}} \right.\] thì \(\left( P \right):x - y + \frac{1}{2}z - 2 = 0\) (loại) vì \(C \in \left( P \right)\).

TH2: \[\left\{ {\begin{array}{*{20}{c}}{a = 2b - 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\\begin{array}{l}b + c - 3 = - 5b - 2c + 3\\\left| {3 + 6b - 6 - b + 2c} \right| = 3\sqrt {{1^2} + {{\left( {2b - 2} \right)}^2} + {b^2}} \end{array}\end{array}} \right.\]\[ \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{a = 2b - 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\\begin{array}{l}2b + c = 2\\\left| {3 + 6b - 6 - b + 2c} \right| = 3\sqrt {{1^2} + {{\left( {2b - 2} \right)}^2} + {b^2}} \end{array}\end{array}} \right.\]

\[ \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{a = 2b - 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\\begin{array}{l}c = 2 - 2b\\\left| {3 + 6b - 6 - b + 4 - 4b} \right| = 3\sqrt {{1^2} + {{\left( {2b - 2} \right)}^2} + {b^2}} \end{array}\end{array}} \right.\]\[ \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{a = 2b - 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\\begin{array}{l}c = 2 - 2b\\\left| {b + 1} \right| = 3\sqrt {5{b^2} - 8b + 5} \end{array}\end{array}} \right.\]

\[ \Leftrightarrow \left\{ \begin{array}{l}a = 2b - 2\\c = 2 - 2b\\44{b^2} - 74b + 44 = 0\end{array} \right.\] (vô nghiệm).

Vậy \(a + b + c = - 4.\)

Cho hình lăng trụ đứng ABC.A′B′C′ có tất cả các cạnh bằng nhau và bằng 4. Góc giữa đường thẳng AA′ và BC′ bằng