Đáp án:
$\begin{array}{l}
a)M\left( {2 + 2t;3 + t} \right)\\
\Rightarrow {d_{M - \Delta '}} = \frac{{\left| {2 + 2t - 3.\left( {3 + t} \right) + 1} \right|}}{{\sqrt {1 + {{\left( { - 3} \right)}^2}} }} = 3\\
\Rightarrow \frac{{\left| {2t - 3t + 2 - 9 + 1} \right|}}{{\sqrt {10} }} = 3\\
\Rightarrow \left| { - t - 6} \right| = 3\sqrt {10} \\
\Rightarrow {t^2} + 12t + 36 = 90\\
\Rightarrow {t^2} + 12t - 54 = 0\\
\Rightarrow t = - 6 \pm 3\sqrt {10} \\
\Rightarrow \left[ \begin{array}{l}
M\left( { - 10 + 6\sqrt {10} ; - 3 + 3\sqrt {10} } \right)\\
M\left( { - 10 - 6\sqrt {10} ; - 3 - 3\sqrt {10} } \right)
\end{array} \right.\\
b)B\left( {2 + 2t;3 + t} \right)\\
\Rightarrow AB = \sqrt {{{\left( {2 + 2t} \right)}^2} + {{\left( {3 + t - 3} \right)}^2}} \\
= \sqrt {4{t^2} + 8t + 4 + {t^2}} \\
= \sqrt {5{t^2} + 8t + 4} \\
= \sqrt {5{{\left( {t + \frac{4}{5}} \right)}^2} + \frac{4}{5}} \ge \sqrt {\frac{4}{5}} \\
\Rightarrow A{B_{\min }} = \frac{2}{{\sqrt 5 }} \Leftrightarrow t = - \frac{4}{5}\\
\Rightarrow B\left( {\frac{2}{5};\frac{{11}}{5}} \right)
\end{array}$
