 Solution

Subtracting the first row from the other two in the determinant givesThis 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 lineConversely, putting three points on any one line makes their displacements linearly dependent, so the determinant vanishes.

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.