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a.
$\left\{ \begin{array}{l}
{d_1}:\left\{ \begin{array}{l}
2x + y + 1 = 0\\
x - y + z - 1 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{\overrightarrow u _1} = \left[ {{{\overrightarrow n }_1}.{{\overrightarrow n }_2}} \right] = [\left ( 2,1,1 \right ).\left ( 1,-1,1 \right )]\\
= (1; - 2; - 3)\\
{M_1}(0; - 1;0) \in {d_1}
\end{array} \right.\\
{d_2}:\left\{ \begin{array}{l}
3x + y - z + 3 = 0\\
2x - y + 1 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{\overrightarrow u _2} = \left[ {{{\overrightarrow {n'} }_1}.{{\overrightarrow {n'} }_2}} \right] =[\left ( 3,1,-1 \right ).\left ( 2,-1,1 \right )]\\
= ( - 1; - 2; - 5)\parallel (1;2;5)\\
{M_2}(0;1;4) \in {d_2}
\end{array} \right.
\end{array} \right.$
$ \Rightarrow \left\{ \begin{array}{l}
{\overrightarrow u _1} = (1; - 2; - 3)\\
{\overrightarrow u _2} = (1;2;5)\\
\overrightarrow {{M_1}{M_2}} = \left( {0;2;4} \right)\parallel (0;1;2)
\end{array} \right. \Rightarrow \left[ {{{\overrightarrow u }_1}.{{\overrightarrow u }_2}} \right].\overrightarrow {{M_1}{M_2}} = 0 \Rightarrow {d_1};{d_2}$ đồng phẳng.
Do (P) đi qua $
{M_1}(0; - 1;0) $ vó có vector pháp tuyến là:
$\begin{array}{l}
{\overrightarrow n _{(P)}} = \left[ {{{\overrightarrow u }_1}.{{\overrightarrow u }_2}} \right] = ( - 4; - 8;4)\parallel (1;2; - 1)\\
\Rightarrow (P):(x - 0) + 2(y + 1) - (z - 0) = 0 \Leftrightarrow (P):x + 2y - z + 2 = 0
\end{array}$
b.
Giả sử (P) cắt 3 trục tọa độ tại A(a;0;0), B(0;b;0),C(0;0;c). Ta có OA=a, OB=b, OC=c và
$\begin{array}{l}
(P):x + 2y - z + 2 = 0 \Leftrightarrow x + 2y - z = - 2 \Leftrightarrow \frac{x}{{ - 2}} + \frac{y}{{ - 1}} + \frac{z}{2} = 1\\
\Rightarrow (a;b;c) = ( - 2; - 1;2) \Rightarrow V = \frac{1}{6}\left| {abc} \right| = \frac{2}{3}(dvtt)
\end{array}$
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