\(\left(d_1\right):3x-y+1=0\Rightarrow\overrightarrow{n_{PT}}=\left(3;-1\right)\)
\(\left(d_1\right)\perp\left(d_3\right)\Rightarrow\left(d_3\right):x+3y+c=0\)
\(\left(d_3\right)\cap\left(d_1\right)=A\left(x_A;y_A\right)\) là nghiệm hpt \(\left\{{}\begin{matrix}3x-y+1=0\\x+3y+c=0\end{matrix}\right.\)
\(\Rightarrow A\left(\dfrac{-3-c}{10};\dfrac{1-3c}{10}\right)\)
\(\left(d_3\right)\cap\left(d_2\right)=C\left(x_C;y_C\right)\) là nghiệm hpt \(\left\{{}\begin{matrix}x-2y+2=0\\x+3y+c=0\end{matrix}\right.\)
\(\Rightarrow C\left(\dfrac{-6-2c}{5};\dfrac{-4+c}{5}\right)\)
\(\left(d_1\right)\cap\left(d_2\right)=B\left(x_B;y_B\right)\) là nghiệm hpt \(\left\{{}\begin{matrix}3x-y+1=0\\x-2y+2=0\end{matrix}\right.\)
\(\Rightarrow B\left(0;1\right)\)
\(AC=\sqrt{\left(\dfrac{-3-c}{10}+\dfrac{6+2c}{5}\right)^2+\left(\dfrac{1-3c}{10}+\dfrac{4-c}{5}\right)^2}\)
\(\Rightarrow AC=\dfrac{1}{10}\sqrt{\left(9+3c\right)^2+\left(9-5c\right)^2}\)
Tương tự \(AB=\dfrac{1}{10}\sqrt{\left(c+3\right)^2+\left(9+3c\right)^2}\)
\(S_{ABC}=\dfrac{1}{2}.AC.AB=5\)
\(\Leftrightarrow\dfrac{1}{2}.\dfrac{1}{10}\sqrt{\left(9+3c\right)^2+\left(9-5c\right)^2}.\dfrac{1}{10}\sqrt{\left(c+3\right)^2+\left(9+3c\right)^2}=5\)
\(\Leftrightarrow\sqrt{\left(9+3c\right)^2+\left(9-5c\right)^2}.\sqrt{\left(c+3\right)^2+\left(9+3c\right)^2}=1000\)
Sau khi giải phương trình trên, ta được \(c=7;c=-13\)
Vậy phương trình tổng quát của \(\left(d_3\right):\left[{}\begin{matrix}x+3y+7=0\\x+3y-13=0\end{matrix}\right.\)
