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I have a question that I could not find on any websites. The question:

Find an equation for the plane that contains the line $x = 1 + t, y = 3t, z = 2t$ and is parallel to the line of intersection of the planes $-x + 2y + z = 0$ and $x + z + 1 = 0$.

I don't know where to start. Can someone give a hint to start me off? Thanks.

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  • $\begingroup$ Welcome to Math SE! . Please read this How to ask a good question to begin with and share your thoughts and efforts in the question $\endgroup$ Nov 30 '19 at 16:17
  • $\begingroup$ That is the second time you said that... and I have read it already. Please stop saying that and I will appreciate it. $\endgroup$
    – Human
    Dec 1 '19 at 11:33
  • $\begingroup$ If you have already read it , then kindly try to ask properly the next time. $\endgroup$ Dec 1 '19 at 11:34
  • $\begingroup$ Got it. I will! $\endgroup$
    – Human
    Dec 1 '19 at 17:41
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Ask yourself what information you need in order to produce an equation of a plane, e.g., three noncolinear points on the plane, a point on the plane and a normal vector, &c. Then, look at the information you’ve been given and try to figure out how to get the data that you need from the given data. For instance, from the line that must lie on the plane you can get two of the three noncolinear points that you need; from two lines that are parallel to the plane (and are not themselves parallel) you can get its normal, and so on.

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  • $\begingroup$ Well said. I now, using the help of this, solved many other questions I previously thought were extremely hard!😉👍👍👍 $\endgroup$
    – Human
    Dec 1 '19 at 16:51
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Calling

$$ l_0\to p =p_0 + t \vec v_0\ \ \cases{p_0 = (1,0,0)\\ \vec v_0 = (1,3,2)}\\ \Pi_1\to (p-p_1)\cdot \vec n_1 = 0\ \ \cases{p_1 = (0,0,0)\\ \vec n_1 = (-1,2,1)}\\ \Pi_2\to (p-p_2)\cdot \vec n_2 = 0\ \ \cases{p_2 = (0,0,-1)\\ \vec n_1 = (1,0,1)}\\ $$

and

$$ \Pi_1\cap\Pi_2 = l_{12} \to p = p_{12}+ t\vec v_{12}\ \ \cases{p_{12}=(0,0,-\frac 12)\\ \vec v_{12}= \vec n_1\times \vec n_2} $$

now if $l_0\cap l_{12} = \emptyset$, the sought plane $\Pi_*$ is such that

$$ \Pi_* \to (p-p_*)\cdot \vec n_*=0\ \ \cases{l_0\in\Pi_*\\ l_{12}\cap \Pi_* = \emptyset} $$

or

$$ (p_0+t\vec v_0-p_*)\cdot\vec n_* = 0\ \ \ \cases{(p_0-p_*)\cdot\vec n_* = 0\\ \vec v_0\cdot \vec n_* = 0}\\ (p_{12}+t\vec v_{12}-p_*)\cdot \vec n_*\ne 0\ \ \ \cases{(p_{12}-p_*)\cdot \vec n_* \ne 0\\ \vec v_{12}\cdot \vec n_* = 0} $$

so choosing

$$ p_* = p_0,\ \ \ \vec n_* = \vec v_0\times\vec v_{12},\ \ \ p_{12}\in l_{12} $$

the plane $\Pi_*$ is fully characterized

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  • $\begingroup$ Sorry, I don't really understand all this notation. I found the other solution clearer, even though it did not tell me an answer. It was also much simpler! $\endgroup$
    – Human
    Dec 1 '19 at 11:01
  • $\begingroup$ Next time, please use less notation. Many people will get confused(like me) $\endgroup$
    – Human
    Dec 1 '19 at 17:41

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