Given a Line Parametrization, Finding another Equation So I am given a line $l$ with the parameterization, $x=t, y=2t, z=3t$. Now let some point, $p$ be a plane that contains the line $l$ and the point $(2,2,2)$. So given this, how do I find an equation for $p$ in the form $ax+by+cz=d$?
My thoughts:
I think that first I have to deal with the $x,y,z$ and find in terms of $t$. I'm not sure though.
 A: One possible approach is the following:
You know that the point $(2,2,2)$ is on the plane $p$, and you know that the line $l$ lies on the plane $p$. In particular, you know that for any value of $t$ you want, you can plug in $t$ into the three parametric equations given and get a point on the line (and hence on the plane). 
If you do this with two points on the line, you will now have three points on the plane (the point (2,2,2), as well as the two points on the line that you found). Note that the point (2,2,2) is not on the line $l$ (why not?).
Now that you have three non-linear points on a plane, then you should be able to answer the question. (Hint, you can find the normal vector to the plane.)
A: in this particular case the point $L_0=(1,1,1)$ lies on the line ($t = 0$). since this point and the point $P=(2,2,2)$ lie in the plane, so does the origin $(0,0,0)$. so its equation is:
$$
ax+by+cz=0 \tag{1}
$$
now any point in the plane is a linear combination $\lambda P + (1-\lambda) L_t$, where $L_t$ is the point on the line with parameter $t$. so, substituting in (1)
$$
a(2\lambda +(1-\lambda)(1+t)) + b(2\lambda +(1-\lambda)(1+2t))+c(2\lambda+(1-\lambda)(1+3t))=0 \tag{2}
$$
since (2) must hold for all values of $t$ and $\lambda$ we have (constant term), 
$$
a+b+c = 0
$$
and (coefficient of $t$)
$$
a+2b+3c=0
$$
this gives $b=-2c$ and $a=c$ so the equation of the plane is:
$$
x-2y+z=0
$$
