Show that the line of intersection of the planes\[x+2y-z=2\mbox{ and } 3x+2y+2z=7\]is parallel to the line \[\langle x,y,x \rangle = \langle 1+6t,3-5t,2-4t \rangle\]

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Show that the line of intersection of the planes\[x+2y-z=2\mbox{ and } 3x+2y+2z=7\]is parallel to the line \[\langle x,y,x \rangle = \langle 1+6t,3-5t,2-4t \rangle\]

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my attempt at a solution. First I found normal vectors to the planes. \[\vec{n}=\langle 1,2,-1 \rangle\]\[\vec{m}=\langle 3,2,2\rangle\]then the cross product to get the vector parallel to that.\[\vec{n}\times \vec{m} = \langle 6, -5, -4 \rangle\]
so wouldn't this be the direction of the line formed by the intersection of the two planes?
so maybe what I am doing wrong is finding the direction of the second line. how would I do this? find two points on the line and then use the displacement vector?

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NICE just figured it out myself, question closed.
nice

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