Let us have 3 vectors in \mathbb R^3: (a_1,b_1,c_1),(a_2,b_2,c_2) and (a_3,b_3,

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 ${\displaystyle (a_2,b_2,c_2)}$ and consider the function ${\displaystyle (a,b,c)\mapsto f(a,b,c,a_2,b_2,c_2)}$. Let's call this function ${\displaystyle gcolon{\mathbb{R}}^3\to \mathbb{R}}$. You want g to be a linear function and that ${\displaystyle g(a,b,c)=0}$ if (a,b,c) linearly depends on ${\displaystyle (a_2,b_2,c_2)}$. However, the set of all vectors which linearly depend on ${\displaystyle (a_2,b_2,c_2)}$ is a line in ${\displaystyle {\mathbb{R}}^3}$ (all scalar multiples of ${\displaystyle (a_2,b_2,c_2)}$) while the zero set of a linear function g is either a plane (if the function is not trivial) or ${\displaystyle {\mathbb{R}}^3}$ (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 ${\displaystyle (a_1,b_1,c_1)}$ and ${\displaystyle (a_2,b_2,c_2)}$ are linearly dependent if and only if the matrix
${\displaystyle A=\left(\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right)}$
has rank ${\displaystyle \le 1}$. This can be checked using determinants. The matrix A has rank ${\displaystyle \le 1}$ iff all ${\displaystyle 2\times 2}$ 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:
${\displaystyle a_1{b}_2-b_1{a}_2=0,a_1{c}_2-c_1{a}_2=0,b_1{c}_2-c_1{b}_2=0}$.
If you want, you can combine them into a single equation
${\displaystyle f(a_1,b_1,c_1,a_2,b_2,c_2)={(a_1{b}_2-b_1{a}_2)}^2+{(a_1{c}_2-c_1{a}_2)}^2+{(b_1{c}_2-c_1{b}_2)}^2=0}$.
However, this f is not linear in ${\displaystyle (a_1,b_1,c_1)}$ (or in ${\displaystyle (a_2,b_2,c_2)}$).
It might be easier to see the argument in the case of one vector in ${\displaystyle {\mathbb{R}}^2}$ (instead of two vectors in ${\displaystyle {\mathbb{R}}^3}$). A single vector (a,b) in ${\displaystyle {\mathbb{R}}^2}$ is "linearly dependent" iff a=b=0 which gives you two scalar linear equations. You can combine them into a single equation ${\displaystyle a_2+b^2=0}$ but this is not linear in (a,b).