Plane intersection by normal line

Question

I have a plane $V$ with cartesian equation $2x - y + 3z = 10$. I also have a line that is perpendicular to the plane $V$ with parametric equation $(x, y, z) = (1, 0, -2) + (2, -1, 3)t.$

How do I find the point of intersection between the line and plane?

Answer 1

Hint: Plug $$x=1+2t$$ $$y=-t$$ $$z=-2+3t$$ in the equation for the plane and compute $$t$$.