Let `a ,b ,c` be real numbers with `a^2+b^2+c^2=1.` Show that the equation `|a x-b y-c b x-a y c x+a b x+a y-a x+b y-cc y+b c x+a c y+b-a x-b y+c|=0` represents a straight line.

Explanation

Given .
` |{:(ax-by-c,,bx+ay,,cx+a),(bx+ay,,-ax +by-c,,cy+b),(cx+a,,cy+b,,-ax-by+c):}|=0`
`" or " |{:((a^(2)+b^(2)+c^(2)),,bx+ay,,cx+a),((a^(2)+b^(2)+c^(2)),,-ax +by-c,,cy+b),((a^(2)+b^(2)+c^(2)),,cy+b,,-ax-by+c):}| =0`
[Applying `C_(1) to aC_(1) +bC_(2)+cC_(3)]`
`" or " |{:(x,,bx+ay,,cx+a),(y,,-ax +by-c,,cy+b),(1,,cy+b,,-ax-by+c):}|=0`
[Applying `C_(2) to C_(2) -bC_(1) " and "C_(3) to C_(3) -cC_(1)]`
`" or " |{:(x,,ay,,a),(y,,-ax-c,,b),(x^(2)+y^(2)+1,,0,,0):}|=0`
[Applying `R_(3) to R_(3) + xR_(1)+yR_(2)` we get
`" or " (x^(2) +y^(2) +1)(aby+a^(2)x+ac) =0 " or "ax+ by + c=0`