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I can do the first part, the trouble I am having is finding the plane. How can I do this?

first, solve the system
\[ \large \begin{cases}
x+2y-z=2\\ 3x+2y+2z=7
\end{cases} \]

they are parallel lines

what are parallel line?

lines*

I need the plane defined by the two lines.

you are given the planes

and you are only given one line

it's supposed to be parallel to the line, I am getting confused

it is. but two parallel lines can still define a plane.

I'll let @helder_edwin take it then, I am a bit confused as to what you're asking

^right, that's what the Q seems to be asking

indeed

why can they not? I see no reason why they can't.

a plane is two-dimensional whareas a line is one-dimensional

you need a point and a vector to define a plane

but the point cannot belong to the line.

right, I mean a point in the plane and a normal vector|dw:1349383884636:dw|

vectors can be moved, so having two does not define any point

but these are not vectors. These are lines!

but you can't get another line because these vectors are linearly dependent

doing that I find that the point (5/2,-1/4,0) is on the intersection of the two planes.

now I have a point and a direction, meaning I have a line. How do I find this line?

ok I have done this and get \[\vec{r}(t)=\langle 5/2+6t, -1/4-5t,-4t \rangle\]

looks reasonable enough

oh I see, so now you want the plane between this line and the given one, eh?

so two lines. need this plane!

yes. and by doing that u got the second direction u needed.

... and you're welcome :)