Đáp án:
$\begin{array}{l}
a)d//{d_1}\\
\Leftrightarrow \left\{ \begin{array}{l}
{m^2} + 2m = m + 6\\
m + 1 \ne - 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{m^2} + m - 6 = 0\\
m \ne - 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left( {m - 2} \right)\left( {m + 3} \right) = 0\\
m \ne - 3
\end{array} \right.\\
\Leftrightarrow m = 2\\
Vậy\,m = 2\\
b)d \bot {d_2}\\
\Leftrightarrow \left( {{m^2} + 2m} \right).\dfrac{{ - 1}}{3} = - 1\\
\Leftrightarrow {m^2} + 2m = 3\\
\Leftrightarrow {m^2} + 2m - 3 = 0\\
\Leftrightarrow \left( {m - 1} \right)\left( {m + 3} \right) = 0\\
\Leftrightarrow m = 1;m = - 3\\
Vậy\,m = 1;m = - 3\\
c)d \equiv {d_3}\\
\Leftrightarrow \left\{ \begin{array}{l}
{m^2} + 2m = - {m^2}\\
m + 1 = 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
2{m^2} + 2m = 0\\
m = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
2m\left( {m + 1} \right) = 0\\
m = 0
\end{array} \right.\\
\Leftrightarrow m = 0\\
Vậy\,m = 0\\
d)Xet:{d_4};{d_5}\\
\Leftrightarrow 2x - 3 = - 3x - 8\\
\Leftrightarrow 5x = - 5\\
\Leftrightarrow x = - 1\\
\Leftrightarrow y = 2x - 3 = 2.\left( { - 1} \right) - 3 = - 5\\
\Leftrightarrow {d_4} \cap {d_5} = \left( { - 1; - 5} \right)\\
\Leftrightarrow \left( { - 1; - 5} \right) \in d\\
\Leftrightarrow - 5 = \left( {{m^2} + 2m} \right).\left( { - 1} \right) + m + 1\\
\Leftrightarrow - {m^2} - 2m + m + 1 = - 5\\
\Leftrightarrow {m^2} + m - 6 = 0\\
\Leftrightarrow \left( {m - 2} \right)\left( {m + 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
m = 2\\
m = - 3
\end{array} \right.\\
Vậy\,m = 2;m = - 3
\end{array}$
