Calculus
posted by Mackenzie .
Find the parametric equations for the line of intersection of the planes x+y+z=3 and xy+2z=2
I took the cross product of the 2 equations and got 3ij2k
I then set z=0 and got x=5/2 and y=1/2.
I got:
x=5/2 +3t
y=1/2t
z=2t
However, the answers are supposed to be:
x=5/2(3/2)t
y=1/2+(1/2)t
z=t
What is my procedure missing to get there?

I think your method is correct.
You took the cross product of the two vectors of the plane <1,1,1>x<1,1,2> and got the direction vector v=<3,1,2>
You combined the two equations and got (0,5/2,1/2), which gives you the position
There are many different solutions for an equation of a line. So don't think that your answer is wrong. 
*oops (5/2,1/2,0) for position
Looking at the answer again,
it looks like they divided the vector by 2 
Do you know why they divided it by 2?