Answer :

Let


Given that Δ = 0.


Recall that the value of a determinant remains same if we apply the operation Ri Ri + kRj or Ci Ci + kCj.


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,


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