Trong không gian cho hai đường thẳng
$(d_1): \frac{x-1}{3}=\frac{y+1}{-1}=\frac{z+1}{2} ;  (d_2):  \begin{cases}x+y-z-2=0 \\ x+3y-12=0 \end{cases}$
Viết phương trình mặt phẳng $(P)$ chứa cả $d_1,d_2$.
Vì $(P)$ chứa $d_2$ nên $(P)$ thuộc "chùm mặt phẳng"
$\alpha (x+y-z-2)+\beta(x+3y-12)=0  (1)  , \alpha^2+\beta^2>0$.
Dễ dàng chứng minh được $d_1//d_2 $ nên $d_1// (P)$
Vì thế $d_1\in (P)$ nếu như $M(1;-2;-1)\in (P)$   (ở đây $  M\in d_1$ ).
Vì $M\in (P)$ nên từ $(1)$ ta có phương trình : $-2\alpha-17\beta=0  (2)$
Từ $(2)$ và do $\alpha^2+\beta^2>0$, nên chọn $\beta=-2; \alpha=17$.
Thay vào $(1)$ ta có: $(P): 15x+11y-17z-10=0$.

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