= Solution
{c}
Subtracting the first row from the other two in the <determinant> gives
$$
f(\mathbf a,\mathbf b,\mathbf c)
=\det\begin{pmatrix}b_1-a_1&b_2-a_2\\c_1-a_1&c_2-a_2\end{pmatrix}.
$$
This vanishes exactly when the two <displacements> $\mathbf b-\mathbf a$ and $\mathbf c-\mathbf a$ are <linearly dependent>. If $\mathbf a\ne\mathbf b$, that means $\mathbf c-\mathbf a=t(\mathbf b-\mathbf a)$ for some <real number> $t$. Thus all three <points> lie on the <line>
$$
\mathbf x=\mathbf a+s(\mathbf b-\mathbf a),\qquad s\in\mathbb R.
$$
Conversely, putting three <points> on any one <line> makes their <displacements> <linearly dependent>, so the <determinant> vanishes.
\b[The equivalence remains true without distinctness.] If exactly two <points> coincide, the two distinct locations determine a <line> containing all three, and two rows of the original <determinant> coincide. If all three <points> coincide, any <line> through that <point> works and the <determinant> is again zero. Existence persists, although uniqueness of the <line> can fail.
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