No, this is not possible. Let's assume we have such an f which satisfies all three of your properties. Fix some non-zero vector and consider the function . Let's call this function . You want g to be a linear function and that if (a,b,c) linearly depends on . However, the set of all vectors which linearly depend on is a line in (all scalar multiples of ) while the zero set of a linear function g is either a plane (if the function is not trivial) or (if the function is identically zero).
However, if you are willing to relax your conditions, you can get what you are looking for. The vectors and are linearly dependent if and only if the matrix
has rank . This can be checked using determinants. The matrix A has rank iff all minors have determinant zero. This gives you three linear equations (not a scalar equation) which give you a sufficient and necessary condition for linear dependance:
If you want, you can combine them into a single equation
However, this f is not linear in (or in ).
It might be easier to see the argument in the case of one vector in (instead of two vectors in ). A single vector (a,b) in is "linearly dependent" iff a=b=0 which gives you two scalar linear equations. You can combine them into a single equation but this is not linear in (a,b).