Finding the parametric equation of a curve

Question

The problem is to find the parametric equation of the line that is tangent to the line of intersection between the plane $x+2y+3z=6$ and the surface $x^2+y^2=2$ and passes through the point $(1,1,1)$.

How would I solve this problem?

Answer 1

Normal vector of the given plane $(1, 2, 3)$
Normal vector of the given surface $(2x, 2y, 0)$ or $(x, y, 0)$ or

$(1, 1, 0) \,$ at point $(1, 1, 1)$ which is on the plane and also on the given surface.

The cross product of the normal vectors $(1, 2, 3)$ and $(1, 1, 0)$ will be the vector of the line tangent to the intersection of both the given plane and the surface which comes to $(3, -3, 1)$.

So parametric equation of the line passing through point $(1, 1, 1)$

is $(1, 1, 1) + t (3, -3, 1)$

Answer 2

If the surfaces are defined by $f(x,y,z)=0$ and $g(x,y,z)=0$, then $\nabla f \times \nabla g$ is a vector tangent to the line of intersection