How to determine the equation of plane which passes through the origin and contains the line x = 2+3t , y = 1-4t, z= 6-t.
I set t = 1 and t = 0 to find the another point, so I can get (0,0,0), (5,-3,5) and (2,1,6)
then do the next step
Is it correct ?
The plane is parallel to $(3,-4,-1)$ and $(2,1,6)-(0,0,0)=(2,1,6)$. So
$$(3,-4,-1)\times(2,1,6)=(23,20,-11)$$
is normal to the plane.
The plane is $23x+20y-11z=0$.