Usng properties of determinants , prove that ∣∣ ∣∣111+3x1+3y1111+3z1∣∣ ∣∣=9(3xyz+xy+yz+zx)
LHS : Let Δ=∣∣ ∣∣111+3x1+3y1111+3z1∣∣ ∣∣By C1→C1−C2,C2→C2−C3⇒Δ=∣∣ ∣∣0−3x1+3x3y01−3z3z1∣∣ ∣∣ Expanding along R1
⇒Δ=9{0+x(y+z)+(1+3x)(yz−0)}
⇒Δ=9{3xyz+xy+yz+zx} =RHS