Question

Let $a$, $b$, $c$ be real numbers with $a^2+b^2+c^2=1$.
show that the equation
$\left|\begin{array}{ccc}ax-by-c& bx+ay& cx+a\\ bx+ay& -ax+by-c& cy+b\\ cx+a& cy+b& -ax-by+c\end{array}\right|=0$
represents a straight line.

Answer 1

Put $z=1$
$\therefore \Delta =(x^2+y^2+1^2)(a^2+b^2+c^2)(ax+by+c)=0$
$\Rightarrow ax+by+c=0$ as $x^2+y^2+1\ne 0$
and $a^2+b^2+c^2=1$ given.
Hence $\Delta =0$ represents the line $ax+by+c=0$.