Question

Find the equation of a line, that lies on a plane `2x-3y+4z=8`, perpendicular to a line `(x-1)/3=(y-1)/2=(z-2)/3` and passes through the point `(1,2,-1)`?

Answer 1

No such line can exist because the point `(1,2,-1)` does not lie on the plane `2x-3y+4z=8`, therefore, the line cannot lie on the plane.

Explanation:

An additional reason why no such line can exist is that the direction of the line is `vecl = 3hati+2hatj+3hatk`; this means that the only planes that can have lines that intersect the given line perpendicularly must be of the form `3x+2y+3z = c` where `c` is any real number. All other planes cannot have lines that intersect the given line perpendicularly.