Đáp án:
$\begin{array}{l}
\Delta //d\\
\Rightarrow \Delta :3x - 4y + c = 0\left( {c \ne - 2} \right)\\
\left( C \right):{x^2} + {y^2} - 6x + 4y - 12 = 0\\
\Rightarrow {\left( {x - 3} \right)^2} + {\left( {y + 2} \right)^2} = 25\\
\Rightarrow Tam\,I\left( {3; - 2} \right);R = 5\\
AB = 8\\
\Rightarrow {d_{I - \Delta }} = \sqrt {{R^2} - {{\left( {\dfrac{{AB}}{2}} \right)}^2}} = 3\\
\Rightarrow \dfrac{{\left| {3.3 - 4.\left( { - 2} \right) + c} \right|}}{{\sqrt {{3^2} + {{\left( { - 4} \right)}^2}} }} = 3\\
\Rightarrow \dfrac{{\left| {c + 17} \right|}}{5} = 3\\
\Rightarrow \left| {c + 17} \right| = 15\\
\Rightarrow \left[ \begin{array}{l}
c = - 2\left( {ktm} \right)\\
c = - 32\left( {tm} \right)
\end{array} \right.\\
\Rightarrow \Delta :3x - 4y - 32 = 0
\end{array}$
