Gọi M, N lần lượt là giao điểm của d và \({{d}_{1}},\,\,{{d}_{2}}\) \(\Rightarrow \)Giả sử \(M(1-t;3t;-2+t),\,\,N(-1+3{{t}_{1}};2+4{{t}_{1}};\,\,1+{{t}_{1}})\)

\(\Rightarrow \overrightarrow{MA}=\left( -5+t;7-3t;7-t \right),\,\,\overrightarrow{MB}=\left( -3-3{{t}_{1}};5-4{{t}_{1}};4-{{t}_{1}} \right)\)

Vì A, M, N thẳng hàng nên \(\frac{-5+t}{-3-3{{t}_{1}}}=\frac{7-3t}{5-4{{t}_{1}}}=\frac{7-t}{4-{{t}_{1}}}\Leftrightarrow \left\{ \begin{align}  & \left( -5+t \right)\left( 5-4{{t}_{1}} \right)=\left( 7-3t \right)\left( -3-3{{t}_{1}} \right) \\ & \left( -5+t \right)\left( 4-{{t}_{1}} \right)=\left( 7-t \right)\left( -3-3{{t}_{1}} \right) \\\end{align} \right.\)

\(\begin{array}{l}
\Leftrightarrow \left\{ \begin{array}{l}
- 25 + 20{t_1} + 5t - 4t{t_1} = - 21 - 21{t_1} + 9t + 9t{t_1}\\
- 20 + 5{t_1} + 4t - t{t_1} = - 21 - 21{t_1} + 3t + 3t{t_1}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
41{t_1} - 4t - 13t{t_1} = 4\\
26{t_1} + t - 4t{t_1} = - 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
41{t_1} - 4t - 13t{t_1} = 4\\
104{t_1} + 4t - 16t{t_1} = - 4
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
145{t_1} - 29t{t_1} = 0\\
26{t_1} + t - 4t{t_1} = - 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
{t_1} = 0\\
t = 5
\end{array} \right.\\
26{t_1} + t - 4t{t_1} = - 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
{t_1} = 0\\
t = - 1
\end{array} \right.\\
\left\{ \begin{array}{l}
t = 5\\
{t_1} = - 1
\end{array} \right.\,\,(L)
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{t_1} = 0\\
t = - 1
\end{array} \right.
\end{array}\)

Phương trình đường thẳng d  đi qua \(M\left( 2;-3;-3 \right)\) và có 1 VTCP \(\overrightarrow{u}=\left( -3;5;4 \right)\) là:   \(\left\{ \begin{align}  & x=2-3t \\ & y=-3+5t \\ & z=-3+4t \\\end{align} \right.\)

Chọn: A