Using the property of determinants and without expanding. ∣∣ ∣∣xax+ayby+bzcz+c∣∣ ∣∣=0
Let A=∣∣ ∣∣xax+ayby+bzcz+c∣∣ ∣∣ Applying C3→C3−C1, we get A=∣∣ ∣∣xaaybbzcc∣∣ ∣∣=0 [Since, the two columns (C2=C3) of the determinants are identical]
Using the property of determinants and without expanding.
∣∣ ∣∣b+cq+ry+xc+ar+pz+xa+bp+qx+y∣∣ ∣∣=2∣∣ ∣∣apxbqycrz∣∣ ∣∣
Using the property of determinants and without expanding. ∣∣ ∣ ∣∣1bca(b+c)1cab(c+a)1abc(a+b)∣∣ ∣ ∣∣=0