Chứng minh $d_{1}//d_{2}$

Lấy điểm $A \in d_{1},B \in d_{2}$, khi đó đường thẳng AB nằm trên mặt phẳng (P)

$\left. \Rightarrow\left\{ \begin{array}{l} {d_{1} \subset (P)} \\ {AB \subset (P)} \end{array} \right.\Rightarrow\left\{ \begin{array}{l} {\overset{\rightarrow}{u_{1}}//(P)} \\ {\overset{\rightarrow}{AB}//(P)} \end{array} \right.\Rightarrow\left\lbrack {\overset{\rightarrow}{u_{1}},\overset{\rightarrow}{AB}} \right\rbrack = \overset{\rightarrow}{n_{(P)}} \right.$

Ta có: $\left. \overset{\rightarrow}{u_{1}} = \left( {2; - 1; - 1} \right),\,\overset{\rightarrow}{u_{2}} = \left( {- 2;1;1} \right)\Rightarrow\overset{\rightarrow}{u_{1}} = - \overset{\rightarrow}{u_{2}}\Rightarrow d_{1}//d_{2} \right.$

Lấy điểm $\left. A\left( {5;1;5} \right) \in d_{1},B\left( {9;0; - 2} \right) \in d_{2}\Rightarrow\overset{\rightarrow}{AB} = \left( {4; - 1; - 7} \right) \right.$

$\left. \Rightarrow\left\lbrack {\overset{\rightarrow}{u_{1}},\overset{\rightarrow}{AB}} \right\rbrack = \left( {6;10;2} \right) = 2\left( {3;5;1} \right) \right.$

Mà $\left. \left\{ \begin{array}{l} {d_{1} \subset (P)} \\ {AB \subset (P)} \end{array} \right.\Rightarrow\left\{ \begin{array}{l} {\overset{\rightarrow}{u_{1}}//(P)} \\ {\overset{\rightarrow}{AB}//(P)} \end{array} \right.\Rightarrow\left\lbrack {\overset{\rightarrow}{u_{1}},\overset{\rightarrow}{AB}} \right\rbrack = \overset{\rightarrow}{n_{(P)}} \right.$

Mặt phẳng đi qua $A\left( {5;1;5} \right)$và có vectơ pháp tuyến ${\overset{\rightarrow}{n}}_{(P)} = \left( {3;5;1} \right)$ có phương trình là: $\left. 3\left( {x - 5} \right) + 5\left( {y - 1} \right) + \left( {z - 5} \right) = 0\Leftrightarrow 3x + 5y + z - 25 = 0 \right.$