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Find parametric equations for the line
$x − 2y + 3z = 4, 2x + y − z = −2$

I know that the first thing you do is cross the vectors perpendicular to the planes, but I'm having a little trouble finding the point of intersection - can I just set $x, y,$ or $z$ to $0$ and take it from there?

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    Wrong question to ask - I'm sleep deprived right now as finals are around the corner. I guess what I was asking in the question was for these two equations, can I just set x, y, or z to 0, and then solving for the two unknowns? I'm terribly confused.2012-12-08

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It seems that you are looking for a point lying on the intersection of two planes which is our wanted line. Take $z=t$ into two equations, you get:

$x-2y+3t=4, 2x+y-t=-2$ Now eliminate the $y$ by every method you know:

$5x+t=0$

Do the same for $x$:

$y=\frac{7}{5}t-2$

So the points are as forms: $\left(\frac{t}{-5},\frac{7}{5}t-2,t\right)$

Take for example $t=0$....

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    Oh man this helps so much.. thanks! I'm gonna do a couple more problems then head to bed, I'm sure I'm just too fatigued to figure things out right now2012-12-08
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Points on the line that is the intersection of the two planes will satisfy the system of linear equations $x-2y+3z=4\\2x+y-z=-2$ which has as it's solution $x=-t/5\\ y=-2+7t/5\\ z=t$ where $t$ is a parameter which maps out the intersection line as it varies over $\mathbb{R}$.

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    @GerryMyerson You are correct.2012-12-08