The direction vector of the normal is the normal vector of the surface at point P. According to the solution of the normal vector, let P(x0, y0, z0) have
k=Fx(x0,y0,z0)
3k=Fy(x0,y0,z0)
4k=Fz(x0,y0,z0)
Here Fx, Fy and Fz represent the partial derivatives of x, y and z.
So we get the equation.
k=2x0-2
3k=6
4k=-2z0
The solution is k = 2, x0 = 2 and z0 =-4. Then substitute it into the surface equation to get y0= 10/3. Write your own line equation.