[Math] How to find plane equation by line and plane that perpendicular to

Question

Find an equation for the plane that is perpendicular to the plane $2x +2y=1$ and passes through the line $x=1-2t,\;\; y=-1+4t,\;\; z=3-2t$
i know how to find an equation by 2 line or 3 point, but it has two different thing with two different type of equation makes me confuse.

Answer 1

The equation of the line can be written as: $$\frac{x-1}{-2}=\frac{y+1}{4}=\frac{z-3}{-2}=t$$ which tells us that one possible point on the desired plan would be $$A:~(-1,+1,-3)$$ and also the leading vector of the line is: $$\vec{w}:=(-2,4,-2)$$ On the other hand, the normal vector of the given plan is $\vec{u}:= (2,2,0)$. Now find the vector $$\vec{n}:= \vec{u}\times\vec{w}$$ to write the equation of the perpendicular plane.