analytic geometry , perpendicular planes and a line

Question

Find a equation to the plane that contain the line $X = (1,0,2) + t(4,1,0)$ and is perpendicular to the plane $A : 3x + y + z = 0$

Answer 1

The desired plane is

$$(1,0,2) + s(3,1,1) + t(4,1,0)$$

It contains the line X (set s=0) and it contains the line

(1,0,2) + s(3,1,1) (set t=0) which is orthogonal to the plane A.

You only have to transform the parameter form into the normal form.

Answer 2

$\nabla A=(3,1,1)$ so our plane is parellel to both the vectors $(4,1,0)$ and $(3,1,1)$. So it is perpendicular to vector $V=(4,1,0)\times (3,1,1)$. then $V=(1,-4,1)$. Our plane also contain the point (1,0,2) so equation of the plane is $$(1,-4,1).(x-1,y,z-2)=0$$ then we get $x-4y+z=3$.