A=<4,2,3> (positional vector of point A), B=<a,b,c>.
A.B=0 if A and B are orthogonal, so 4a+2b+3c=0, which has many solutions. Imagine a line to represent A, then there are many lines which could be at right angles to it in 3-space.
|B|=√(a2+b2+c2) is the magnitude of B. Its unit vector is B/|B|=<a,b,c>/√(a2+b2+c2).
b=-(4a+3c)/2, so the unit vector is <a,-(4a+3c)/2,c>/√(a2+(4a+3c)2/4+c2), which depends on values of a and c. If a=0 and c=4, for example, the unit vector would be <0,-6,4>/(2√13)=<0,-3/√13,2√13).
If a=1, c=2, then b=-5, and the unit vector would be <1,-5,2>/√30.
There is no constant unit vector. Three equations are needed to determine a, b, c, and we can derive only one from the given information.