The simplest scalar equation to determine whether vectors in $\mathbb R^3$ are linearly dependent Let us have $3$ vectors in $\mathbb R^3$:
$(a_1,b_1,c_1)$, $(a_2,b_2,c_2)$ and $(a_3,b_3,c_3)$
These vectors are linearly dependent if and only if the following scalar equation holds:
$a_1(b_2c_3 - b_3c_2) - b_1(a_2c_3 - a_3c_2) + c_1(a_2b_3 - a_3b_2) = 0$

*

*This equation is a necessary and sufficient condition of lineary dependence.

*This criterion is a scalar equation (not a pair of equations; not a vector form).

*This equation is a linear equation as to $(a_i,b_i,c_i)$ for all $i\in\{1,2,3\}$.

Now let us have $2$ strings in $\mathbb R^3$:
$(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$
Is there a similar way to determine whether this set of vectors in $\mathbb R^3$ is linearly dependent?
I.e. the way such as the following scalar equation:
$f(a_1, b_1, c_1, a_2, b_2, c_2) = 0$
so, that

*

*This equation is a necessary and sufficient condition of linear
dependence.

*This criterion is a scalar equation (not a pair of equations; not a vector form).

*This equation is a linear equation as to $(a_i,b_i,c_i)$ for all $i\in\{1,2\}$.

 A: 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 $g \colon \mathbb{R}^3 \rightarrow \mathbb{R}$. You want $g$ to be a linear function and that $g(a,b,c) = 0$ iff $(a,b,c)$ 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
$$ A = \begin{pmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{pmatrix} $$
has rank $\leq 1$. This can be checked using determinants. The matrix $A$ has rank $\leq 1$ 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:
$$ a_1 b_2 - b_1 a_2 = 0, \,\,\, a_1 c_2 - c_1 a_2 = 0, \,\,\, b_1 c_2 - c_1 b_2 = 0. $$
If you want, you can combine them into a single equation
$$ f(a_1,b_1,c_1,a_2,b_2,c_2) = (a_1 b_2 - b_1 a_2)^2 + (a_1 c_2 - c_1 a_2)^2 + (b_1 c_2 - c_1 b_2)^2 = 0. $$
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 $(a,b)$.
