As the given system of equations has a non-trivial solution. Δ=|
p
a
a
b
q
b
c
c
r
|=0 Applying C2→C2−C1 and C3→C3−C1 Δ=|
p
a−p
a−p
b
q−b
0
c
0
r−c
|=0 Expanding along C3, we get (a−p)|
b
q−b
c
0
|+(r−c)|
p
a−p
b
q−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