Đáp án:
$\begin{array}{l}
{x^2} - 4x + 3 = mx + 3\\
\Rightarrow {x^2} - \left( {m + 4} \right)x = 0\\
\Rightarrow x\left( {x - m - 4} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
x = m + 4\left( {m + 4 \ne 0 \Rightarrow m \ne - 4} \right)
\end{array} \right.\\
\Rightarrow A\left( {0;3} \right);B\left( {m + 4;m\left( {m + 4} \right) + 3} \right)\\
\Rightarrow B\left( {m + 4;{m^2} + 4m + 3} \right)\\
\Rightarrow AB = \sqrt {{{\left( {m + 4} \right)}^2} + {{\left( {{m^2} + 4m} \right)}^2}} \\
= \sqrt {{{\left( {m + 4} \right)}^2} + {m^2}{{\left( {m + 4} \right)}^2}} \\
= \sqrt {{{\left( {m + 4} \right)}^2}\left( {{m^2} + 1} \right)} \\
h = {d_{O - AB}} = {d_{O - y = mx + 3}} = \dfrac{{\left| 3 \right|}}{{\sqrt {{m^2} + 1} }} = \dfrac{3}{{\sqrt {{m^2} + 1} }}\\
\Rightarrow {S_{OAB}} = \dfrac{9}{2}\\
\Rightarrow \dfrac{1}{2}.h.BA = \dfrac{9}{2}\\
\Rightarrow \dfrac{3}{{\sqrt {{m^2} + 1} }}.\sqrt {{{\left( {m + 4} \right)}^2}\left( {{m^2} + 1} \right)} = 9\\
\Rightarrow \sqrt {{{\left( {m + 4} \right)}^2}} = 3\\
\Rightarrow \left| {m + 4} \right| = 3\\
\Rightarrow \left[ \begin{array}{l}
m + 4 = 3\\
m + 4 = - 3
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
m = - 1\left( {tm} \right)\\
m = - 7\left( {tm} \right)
\end{array} \right.\\
\Rightarrow S = - 1 - 7 = - 8
\end{array}$
