Normal vector (4x, 6y, 2z)

At point m (1,-1, 2), n 1 =(2, -3, 2).

3x^2+y^2-z^2 = 0

Normal vector (6x, 2y, -2z)

At point M( 1,-1, 2), n2 =(3,-1, 2).

Tangent direction vector t = n 1 × n2 = (8,10,7)

The tangent equation (x-1)/8 = (y+1)10 = (z-2)/7.

Normal plane equation 8 (x-1)+10 (y+1)+7 (z-2) = 0.

That is 8x+ 10y+7z = 12.

According to the expression of space curve, there are two solutions:

1. Parameter curve form: Find the reciprocal of x, y and z to parameter t respectively, bring in the value of this point to get the tangent vector of this point, and write the tangent and normal plane according to the point direction formula and point method.

2. The form of intersection of two planes: according to the equation, find the partial derivatives of z to x and y to x, then write the tangent vector, and then write the tangent plane and normal plane.