linear equation of a plane from parametric equation

Question

How to find a linear equation of a plane that passes through the point $(6,0,-2)$ and contains the line with parametric equations $x=4-2t$, $y=3+5t$, $z=7+4t$?

Answer 1

Three points in the plane are $(6,0,-2)$, $(4,3, 7$) [on the line when $t=0$],

and $(2,8,11)$ [on the line when $t=1$].

Their differences are vectors in the plane:

$(6,0,-2)-(4,3,7)=(2,-3,-9)$ and $(6,0,-2)-(2,8,11)=(4,-8,-13)$.

The cross-product of the two vectors $(2,-3,-9)\times(4,-8,-13)=(-33,-10,-4)$

is normal to the plane, and so is $(33,10,4)$. So an equation of the plane is $33x+10y+4z=C$,

and you can figure out $C$ using any of the three points we know in the plane. (I got $C=190$.)