Giả sử $\Delta$ có vtcp $\overrightarrow{{{u}_{\Delta }}}=(a;b;c),{{a}^{2}}+{{b}^{2}}+{{c}^{2}}>0$

$\Delta \bot {{d}_{1}}\Leftrightarrow \overrightarrow{{{u}_{\Delta }}}.\overrightarrow{{{u}_{1}}}=0\Leftrightarrow a-b+c=0(1)$

$\left( \Delta ,{{d}_{2}} \right)={{60}^{o}}\Leftrightarrow \cos {{60}^{o}}=\frac{\left| a-b-2c \right|}{\sqrt{1+1+4}.\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}\Leftrightarrow 2{{(a-b-2c)}^{2}}=3({{a}^{2}}+{{b}^{2}}+{{c}^{2}})(2)$

Từ (1) $\Rightarrow b=a+c$ thay vào (2) ta được: $18{{c}^{2}}=3\left[ {{a}^{2}}+{{(a+c)}^{2}}+{{c}^{2}} \right]\Leftrightarrow {{a}^{2}}+ac-2{{c}^{2}}=0$

$\Leftrightarrow (a-c)(a+2c)=0\Rightarrow a=c\vee a=-2c$

• $a=c\Rightarrow b=2c$ chọn $c=1\Rightarrow \overrightarrow{{{u}_{\Delta }}}=(1;2;1)$

Ta có: $\Delta :\frac{x+1}{1}=\frac{y-2}{2}=\frac{z}{1}$

$a=-2c\Rightarrow b=-c$ chọn $c=-1\Rightarrow \overrightarrow{{{u}_{\Delta }}}=(2;1;-1)$

Ta có: $\Delta :\frac{x+1}{2}=\frac{y-2}{1}=\frac{z}{-1}$