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 $({a}_{2},{b}_{2},{c}_{2})$ and consider the function $(a,b,c)\mapsto f(a,b,c,{a}_{2},{b}_{2},{c}_{2})$. Let's call this function $gcolon{\mathbb{R}}^{3}\to \mathbb{R}$. You want g to be a linear function and that $g(a,b,c)=0$ if (a,b,c) linearly depends on $({a}_{2},{b}_{2},{c}_{2})$. However, the set of all vectors which linearly depend on $({a}_{2},{b}_{2},{c}_{2})$ is a line in $\mathbb{R}}^{3$ (all scalar multiples of $({a}_{2},{b}_{2},{c}_{2})$) while the zero set of a linear function g is either a plane (if the function is not trivial) or $\mathbb{R}}^{3$ (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 $({a}_{1},{b}_{1},{c}_{1})$ and $({a}_{2},{b}_{2},{c}_{2})$ are linearly dependent if and only if the matrix

$A=\left(\begin{array}{ccc}{a}_{1}& {b}_{1}& {c}_{1}\\ {a}_{2}& {b}_{2}& {c}_{2}\end{array}\right)$

has rank $\le 1$. This can be checked using determinants. The matrix A has rank $\le 1$ iff all $2\times 2$ 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:

${a}_{1}{b}_{2}-{b}_{1}{a}_{2}=0,\text{}\text{}{a}_{1}{c}_{2}-{c}_{1}{a}_{2}=0,\text{}\text{}{b}_{1}{c}_{2}-{c}_{1}{b}_{2}=0$.

If you want, you can combine them into a single equation

$f({a}_{1},{b}_{1},{c}_{1},{a}_{2},{b}_{2},{c}_{2})={({a}_{1}{b}_{2}-{b}_{1}{a}_{2})}^{2}+{({a}_{1}{c}_{2}-{c}_{1}{a}_{2})}^{2}+{({b}_{1}{c}_{2}-{c}_{1}{b}_{2})}^{2}=0$.

However, this f is not linear in $({a}_{1},{b}_{1},{c}_{1})$ (or in $({a}_{2},{b}_{2},{c}_{2})$).

It might be easier to see the argument in the case of one vector in $\mathbb{R}}^{2$ (instead of two vectors in $\mathbb{R}}^{3$). A single vector (a,b) in $\mathbb{R}}^{2$ is "linearly dependent" iff a=b=0 which gives you two scalar linear equations. You can combine them into a single equation ${a}_{2}+{b}^{2}=0$ but this is not linear in (a,b).