Câu hỏi 7

Gọi \(S\left( {a;b;c} \right).\)

Vì \(S \in \left( P \right) \Rightarrow 2a - 6b - 4c + 7 = 0\,\,\,\left( 1 \right)\)

Theo bài ra ta có:

\(\begin{array}{l}SA = SB = SC\\ \Rightarrow \left\{ \begin{array}{l}SA = SB\\SA = SC\end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l}{\left( {a - 2} \right)^2} + {\left( {b - 4} \right)^2} + {\left( {c + 1} \right)^2} = {\left( {a - 1} \right)^2} + {\left( {b - 4} \right)^2} + {\left( {c + 1} \right)^2}\\{\left( {a - 2} \right)^2} + {\left( {b - 4} \right)^2} + {\left( {c + 1} \right)^2} = {\left( {a - 2} \right)^2} + {\left( {b - 4} \right)^2} + {\left( {c - 3} \right)^2}\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} - 4a - 8b + 2c + 21 =  - 2a - 8b + 2c + 18\\ - 4b - 8b + 2c + 21 =  - 4a - 8b - 6c + 29\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} - 2a =  - 3\\8c = 8\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = \dfrac{3}{2}\\c = 1\end{array} \right.\end{array}\)

Thay vào (1) ta có: \(2.\dfrac{3}{2} - 6b - 4.1 + 7 = 0 \Leftrightarrow b = 1.\)

Khi đó ta có: \(S\left( {\dfrac{3}{2};1;1} \right) \Rightarrow SA = \sqrt {\dfrac{1}{4} + 9 + 4}  = \dfrac{{\sqrt {53} }}{2}\).

Vậy \(l = SA + SB = 2SA = \sqrt {53} .\)

Chọn A.