Prove that the determinant represents a straight line. Let $a,b,c$ be real numbers such that $a^2+b^2+c^2=1$, then show that the equation

represents a straight line.
Expansion of determinant is tedious. I can see that matrix is symmetric but not able to use this fact. Could someone help me with this?
 A: Notice that the inner matrix can be written as
$$T = \mathrm{a}\mathrm{x}^T + \mathrm{x}\mathrm{a}^T - \mathrm{a}^T\mathrm{x},$$
where $\mathrm{a} = (a, b, c)$ and $\mathrm{x} = (x, y, 1)$ are column vectors. This gives
$$T\mathrm{a} = \mathrm{x}, \qquad T\mathrm{x} = \|\mathrm{x}\|^2 \mathrm{a}, \qquad T\mathrm{v} = -\langle \mathrm{a}, \mathrm{x} \rangle \mathrm{v} \tag{*}$$
whenever $\mathrm{v}$ is perpendicular to both $\mathrm{a}$ and $\mathrm{x}$. So if $\mathrm{a}$ and $\mathrm{x}$ are linearly independent, $\text{(*)}$ completely determines the representation of $T$ w.r.t. the basis $\{\mathrm{a}, \mathrm{x}, \mathrm{a}\times\mathrm{x}\}$, from which we compute as
$$\det T = \|\mathrm{x}\|^2 \langle a, x\rangle.$$
By appealing to the continuity or by a similar idea as before, we can check that this relation continues to hold when they are dependent. Now the conclusion can be easily drawn from this formula.
A: \begin{eqnarray*}
 Det\left[
\begin{array}{ccc}
ax-by-c&bx+ay&cx+a \\
bx+ay&-ax+by-c&cy+b\\
cx+a&cy+b&-ax-by+c \\
\end{array}
\right]=(x^2+y^2+1)(a^2+b^2+c^2)(ax+by+c).
\end{eqnarray*}
The first two factors are positive so $\color{red}{ax+by+c=0}$.
