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This is how I solved this problem:

I left $\vec{p}(t)$ in terms of $t$ and changed the parameter of $\vec{q}$ to $s$. I then set the corresponding components of each equal to each other. First the x terms: $$1+3(t) = 3-s$$ Then solve for s $$s=-3t+2$$ Now the y terms: $$2+4t = -4+3s$$ Here I plugged in my s value: $$2+4t=-4+3(-3t+2)$$ And now everything is in terms of t, so solve for t: $$2+4t=-4-9t+6$$ $$4t=9t$$ $$t=0$$ Now I use this t value to solve my original x term equality: $$1+3(0)=3-s$$ Solve for s: $$s=2$$ Now, go back to the original paths and plug in our values, this is the point that the paths will cross: $$\vec{p}(0) = \langle1+0, 2+0\rangle$$ $$=\langle1,2\rangle$$ $$\vec{q}(2) = \langle3-2,-4+3(2)\rangle$$ $$=\langle1,2\rangle$$ Point of intersenction: $\langle1,2\rangle$

As far as part b, do the motions collide? I answered no, because the paths cross this point at different time values.