$\begin{array}{l} (d')\left\{ \begin{array}{l} x = 3 + mt\\ y = 2 + t \end{array} \right. \to \overrightarrow {{u_{d'}}} = \left( {m;1} \right) \to \overrightarrow {{n_{d'}}} = \left( { - 1;m} \right)\\ (d):x + y + 1 = 0 \to \overrightarrow {{n_{d'}}} = \left( {1;1} \right)\\ \cos \left( {d,d'} \right) = \dfrac{{\left| { - 1.1 + m.1} \right|}}{{\sqrt {{1^2} + {{\left( { - 1} \right)}^2}} .\sqrt {{{\left( { - 1} \right)}^2} + {m^2}} }} = \dfrac{{\sqrt 3 }}{2}\\ \Leftrightarrow \left| {m - 1} \right| = \dfrac{{\sqrt 6 }}{2}\sqrt {{m^2} + 1} \\ \Leftrightarrow {\left( {m - 1} \right)^2} = \dfrac{3}{2}\left( {{m^2} + 1} \right)\\ \Leftrightarrow \dfrac{1}{2}{m^2} + 2m + \dfrac{1}{2} = 0\\ \Leftrightarrow {m^2} + 4m + 1 = 0 \Leftrightarrow {m_1} + {m_2} = \dfrac{{ - b}}{a} = - 4 \to B \end{array}$
