How to treat the normal vector of direction vector 2X+4Y+3Z=3 in college mathematics?

2X+4Y+3Z=3 in space geometry represents a plane.

The normal vector of the direction vector refers to the vector perpendicular to this plane, and there are countless vectors.

On this plane, take any three points that are not * * lines, such as A (0 0,0, 1) B (1,-1) c (2,0 1, 1).

Then vector AB=( 1, 1, -2), vector AC=(2,-1, 0).

Try to find the vector as (a, b, c)

Then (a, b, c) is perpendicular to vector AB=( 1, 1, -2) and vector AC=(2,-1, 0).

A+b-2c=0, 2a-b=0.

Let a= 1, substitute b = 2 and c = 3/2.

Therefore, a normal vector of this plane is (1, 2,3/2).

All normal vectors of this plane are t( 1, 2,3/2), where t is a non-zero real number.