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The plane passes through $(3,2,-1)$ and $(1,-1,2)$ and is parallel to the line $v=(1,-1,0) + t(3,2,-2)$

I know how to do this with a vector perpendicular to the plane and that passes through a point, but I'm clueless on this one

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  • $\begingroup$ Find a vector that is perpendicular to the given line, and you shall be able to proceed further. $\endgroup$ – Workaholic Feb 3 '16 at 20:53
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Hint: Use the two points in the plane to get a second direction that lies in the plane, in addition to the direction from the line. Then write down a third vector that's perpendicular to these two (perhaps by considering the cross product).

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  • $\begingroup$ I'm not sure if this is right, but i made a vector from (3,2,-1) to (1,-1,2) to make <-2,-3,3>. Since the parametric equation is parallel, i took its vector, <3,2,-2> and took their cross product to find the normal vector <0,5,5>. Since I went from the point (3,2,-1), I used this as the point to calculate the equation of the plane which I have as: 5y+5z=5 I'm not sure if this is correct though. $\endgroup$ – Joe Jackson Feb 4 '16 at 10:25

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