If ab−cc+ba+cbc−aa−ba+bc=0, the line ax+by+c=0 passes through the fixed point which is

Given, ab−cc+ba+cbc−aa−ba+bc=0⇒ 1aa2b−cc+ba2+acbc−aa2−aba+bc=0Applying C1→C1+bC2+cC3, then⇒ 1aa2+b2+c2b−cc+ba2+b2+c2bc−aa2+b2+c2a+bc=0Applying R2→R2−R1 and R3→R3−R1, then⇒1aa2+b2+c2⋯b−c⋯c−b⋮    0 c −b−a⋮    0 a+c −b=0Expanding along C1, then ⇒    a2+b2+c2ac−b−aa+c−b=0⇒    a2+b2+c2a[(−bc+(b+a)(a+c)]=0⇒    a2+b2+c2−bc+ab+bc+a2+aca=0⇒    a2+b2+c2(a+b+c)=0∵    a2+b2+c2≠0∴    a+b+c=0Therefore, line ax+by+c=0 passes through the fixed point (1, 1).