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.

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:
$$
x-2y+z=0
$$
