The normal is parallel to the known straight line, so the direction vector of the normal is (1, 3,4). Find another point p, and the normal equation can be solved.

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.