In fact, a plane has countless normal vectors, and these normal vectors are all parallel.

Any plane: ax+by+cz+d=0. Let a set of numbers x0, y0 and z0 satisfy the equation, then:

Ax0+by0+cz0+d=0, two expressions are subtracted to get: a(x-x0)+b(y-y0)+c(z-z0)=0, which is the point equation of the plane.

Represents a plane passing through points (x0, y0, z0) and taking n=(a, b, c) as the normal. Ax+by+cz+d=0 is a general equation of plane.

Remember: the coefficients of x, y and z in the equation are a normal vector of the plane.

Your equation is like this, so a normal vector of the plane: n=( 1, 3,2), but this is not unique.

For example, 3n=(3, 9, 6).

The normal vector of plane 2x-y+6=0 is (2,-1, 6). Let the normal vector of the plane be (a, b, c). Because the two planes are perpendicular, the two normal vectors are perpendicular, that is, (2,-1, 6) (a, b, c) = 2a-b+. y = 0.5t+5； Z=-2.5t- 13。

Specific steps of plane normal vector: (undetermined coefficient method) 1, establish a proper rectangular coordinate system 2, set the plane normal vector n=(x, y, z) 3, and find two vectors with straight lines except * * * in the plane, and record them as a=(a 1, A2, a2, a3) b = ..

If it is high school mathematics, you can try to find out the vector BA=( 1, 0,-1), the vector BC=(0, 1, 1) and the vector p=(a, y, z) p perpendicular to BA and BC, and X-Z = 0. ...