How to solve an equation with two lines perpendicular to a plane

Question

Find a plane passing through the line (x − 2)/2 = z + 3, y = 1 and perpendicular to the plane x - 2y + z = 2. How would I go about solving this problem? Would I use two points on the two lines (2,0,-3) and (0,1,0) to create a vector: (-2,1,3) and then find the cross product of this vector and the normal of the plane (1, -2, 1)?

Answer 1

For any real $t$ we see that $(2+2t,1,-3+t)$ is a point, which is placed on the line.

Let $\vec{n}(a,b,c)$ be a vector normal of the needed plane.

Thus, $$a-2b+c=0$$ and $$2a+c=0,$$ which gives $$a=-2b$$ and $$c=4b,$$ which for $b=-1$ gives $$\vec{n}(2,-1,-4)$$ and since the point $(1,1,-3)$ is placed on the needed plane, we obtain the following equation. $$2(x-2)-(y-1)-4(z+3)=0$$ or $$2x-y-4z-15=0.$$