The given is
Line $(1):$ $$x= 1+2t, \space y=4t, \space z= 3-3t$$ Line $(2):$ $$x= 2+t , y=-1-t , z=-3t$$
First I changed the variable in line $2$ to $s$ as to find the point of intersection of the $y$ coord. $$y=4t=-1-s$$ $$s=-1-4t$$ Now we set $t=0$ to solve for $x$ and $z$ $$x=2+(-1-4(0))$$ $$x=1$$ $$z=-3(-1-4(0))$$ $$z=3$$ So our point of intersection is $(1,0,3)$ The $y$ coord being zero means that the intersection is parallel to the $xy$ and $zy$ planes. So now we extrapolate out two vector from our given parametric equations $$\vec V_1= \langle 2,4,-3\rangle$$ $$\vec V_2= \langle 1,-1,-3\rangle$$ $$\vec V_1 X \vec V_2=\langle-15,3,-6\rangle$$ So the equation of the line is $$-15(x-1)+3(y)-6(z-3)=0$$ Am I correct?