Let

Given that Δ = 0.

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying R₁→ R₁ – R₂, we get

Applying R₂→ R₂ – R₃, we get

Expanding the determinant along R₁, we have

Δ = (p – a)[(q – b)(r) – (b)(c – r)] – (b – q)[–a(c – r)]

⇒ Δ = r(p – a)(q – b) – b(p – a)(c – r) + a(b – q)(c – r)

∴ Δ = r(p – a)(q – b) + b(p – a)(r – c) + a(q – b)(r – c)

We have Δ = 0

⇒ r(p – a)(q – b) + b(p – a)(r – c) + a(q – b)(r – c) = 0

On dividing the equation with (p – a)(q – b)(r – c), we get

Thus,