Struggling to begin answering the following question:
Let $L$ be the line given by $x = 3-t, y= 2+t, z = -4+2t$. $L$ intersects the plane $3x-2y+z=1$ at the point $P = (3,2,-4)$. Find parametric equations for the line through $P$ which lies on plane and is perpendicular to $L$.
So far, I know that I need to find some line represented by a vector $n$ which is orthogonal to $L$. So, with the vector of $L$ represented by $v$, I have:
$$n\cdot v = 0 \Rightarrow [a, b, c] \cdot [-1, 1, 2] = 0 \Rightarrow -a + b + 2c = 0$$
I am not sure how to proceed from here, or if I am even on the right track.
You know that the line you want is perpendicular to the line L, which has direction vector $\langle -1,1,2\rangle$, and that the line you want lies in the given plane, which has normal vector $\langle 3, -2, 1\rangle$. So the line you want is orthogonal to both $\langle -1,1,2\rangle$ and $\langle 3, -2, 1\rangle$ and you can use the cross product of these two vectors as the direction vector of the line you want.