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$$ a = \begin{cases} x=-2+s \\ y=2+s \\ z=1-s \end{cases} $$ $$ b = \begin{cases} x=2p \\ y=4+2p \\ z=4+3p \end{cases} $$ Find the equasion of the plane in which a and b cross

So first I thought I can just get the vectors of the system and make a system like this: $$v_{a}=(1,1,-1) $$ $$ v_{b}=(2,2,3) $$ and after that make a matrix: $$[ \begin{matrix} x-1 & y-1 & z+1 \\ 2-1 & 2-1 & 3+1 \\ \end{matrix} ]$$

But I can't do anything with that after that I thought of a thing we did in class which I did not quite understand:

$$ \lambda (-2+s,2+s,1-s)+\mu (2p,4+2p,4+3p)=0 $$

But that also doesn't lead me to anywhere what do I have to do in this one ?

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    What are the `coordinates` of a plane? – 2017-01-22
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    no information given about the plane, we have to find the coordinates ourselfves, – 2017-01-22
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    You don't answer my question: what are coordinates, for a *plane*? – 2017-01-22
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    as I said, no information is given, the whole problem is to find the coordinates of the plane – 2017-01-22

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at first we compute the intersection Point of the two given straight lines solve the System $$2p=-2+s$$ $$4+2p=2+s$$ $$4+3p=1-s$$ from here we get $$p=-1$$ and $$s=0$$ thus we obtain the equation in the form $$[x,y,z]=[-2,2,1]+s[1,1,-1]+t[2,2,3]$$ with the real numbers $$s,t$$

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    what are the coordinates of the plane and from where did s and t come from ? – 2017-01-23