Lời giải:Gọi \(\left\{ \begin{align}
& A=\Delta \cap {{d}_{1}}\Rightarrow A\left( a;-1+2a;a \right) \\
& B=\Delta \cap {{d}_{2}}\Rightarrow B\left( b;1-2b;1+3b \right) \\
\end{align} \right.\Rightarrow \overrightarrow{AB}=\left( -a+b;-2a-2b+2;-a+3b+1 \right).\)
Ta có: \(\overrightarrow{AB}\text{//}{{\vec{u}}_{d}}\Rightarrow \frac{-a+b}{1}=\frac{-2a-2b+2}{4}=\frac{-a+3b+1}{-2}\Rightarrow \left\{ \begin{align}
& -2a+6b=2 \\
& 3a-5b=1 \\
\end{align} \right.\)
\(\Rightarrow \left\{ \begin{align}
& a=2 \\
& b=1 \\
\end{align} \right.\Rightarrow A\left( 2;3;2 \right),B\left( 1;-1;4 \right).\)
\(\Rightarrow \Delta \)qua \(B\left( 1;-1;4 \right)\) và có vectơ chỉ phương là \(\vec{u}=\left( 1;4;-2 \right)\)
\(\Rightarrow \left( \Delta\right):\left\{ \begin{align}
& x=1+t \\
& y=-1+4t \\
& z=4-2t \\
\end{align} \right.\) đi qua điểm \(N\left( 0;-5;6 \right).\)
