Đáp án:
$\begin{array}{l}
a)VTCP:\overrightarrow u = \left( {5; - 1} \right)\\
\Rightarrow {d_1}:5x - y + c = 0\\
Do:A\left( { - 2;3} \right) \in {d_1}\\
\Rightarrow 5.\left( { - 2} \right) - 3 + c = 0\\
\Rightarrow c = 13\\
\Rightarrow {d_1}:5x - y + 13 = 0\\
b)VTPT:\overrightarrow n = \left( {2;5} \right)\\
\Rightarrow VTCP:\overrightarrow m = \left( {5; - 2} \right)\\
\Rightarrow {d_2}:5x - 2y + d = 0\\
B\left( {3; - 1} \right) \in {d_2}\\
\Rightarrow 5.3 - 2.\left( { - 1} \right) + d = 0\\
\Rightarrow d = - 17\\
\Rightarrow {d_2}:5x - 2y - 17 = 0\\
c)M\left( {3;5} \right) \to {d_1}\\
{d_{M - {d_1}}} = \dfrac{{\left| {5.3 - 5 + 13} \right|}}{{\sqrt {{5^2} + {{\left( { - 1} \right)}^2}} }} = \dfrac{{23}}{{\sqrt {26} }} = \dfrac{{23\sqrt {26} }}{{26}}\\
d)\left\{ \begin{array}{l}
{d_1}:5x - y + 13 = 0\\
{d_2}:5x - 2y - 17 = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
5x - y = - 13\\
5x - 2y = 17
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = - \dfrac{{43}}{5}\\
y = - 30
\end{array} \right.\\
\Rightarrow {d_1} \cap {d_2} = \left( { - \dfrac{{43}}{5}; - 30} \right)
\end{array}$
