${\Delta _1}$ đi qua M(0;2;0), $\overrightarrow {{u_1}} = (1; - 1;1)$
${\Delta _2}$ có VTCP $\overrightarrow {{u_2}} = (2;1; - 1)$
(P) qua M, có VTPT $\overrightarrow n \bot \overrightarrow {{u_1}} $ và $\cos \left( {\overrightarrow n ,\overrightarrow {{u_2}} } \right) = \dfrac{1}{2}$
Đặt $\overrightarrow n = (A;B;C)$
Ta có:
$\begin{array}{l}
\left\{ \begin{array}{l}
A - B + C = 0\\
\dfrac{{\left| {2A + B - C} \right|}}{{\sqrt {6({A^2} + {B^2} + {C^2})} }} = \dfrac{1}{2}
\end{array} \right. \to \left\{ \begin{array}{l}
B = A + C\\
6\left| A \right| = \sqrt {6({A^2} + {C^2} + {{\left( {A + C} \right)}^2}}
\end{array} \right. \to \left\{ \begin{array}{l}
B = A + C(1)\\
2{A^2} - AC - {C^2} = 0(2)
\end{array} \right.\\
(2) \to \left( {A - C} \right)\left( {2A + C} \right) = 0\\
\to \left[ \begin{array}{l}
A = C\\
2A = - C
\end{array} \right.
\end{array}$
TH1: A = C. Chọn A = C = 1
Suy ra PT $(P): x + 2y + z - 4 = 0$
TH2: 2A = -C. Chọn A = 1, C = -2
Suy ra PT $(P): x - y - 2z + 2 = 0$