Let a,b,c be real numbers with `a^2 + b^2 + c^2 =1`. Show that the equation `|[ax-by-c,bx+ay,cx+a],[bx+ay,-ax+by-c,cy+b],[cx+a,cy+b,-ax-by+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`