As the given system of equations has a non-trivial solution. Δ=pbcaqcabr=0 Applying C2→C2−C1 and C3→C3−C1Δ=pbca−pq−b0a−p0r−c=0 Expanding along C3, we get (a−p)bcq−b0+(r−c)pba−pq−b=0⇒(a−p)(−c)(q−b)+(r−c){p(q−b)−b(a−p)}=0⇒(p−a)(q−b)c+p(r−c)(q−b)+b(r−c)(p−a)=0 Dividing by (p−a)(q−b)(r−c), we get r−cc+p−ap+q−bb=0⇒p−ap+q−bq+r−cr=q−bq−b+r−cr−c=2