If ∣∣ ∣ ∣∣xnxx+2xx+4ynyn+2yn+4znzn+2zn+4∣∣ ∣ ∣∣=(1y2−1x2)(1z2−1y2)(1x2−1z2) then n=
-4
Δ=∣∣
∣
∣∣xnxx+2xx+4ynyn+2yn+4znzn+2zn+4∣∣
∣
∣∣=xnynzn∣∣
∣
∣∣1x2x41y2y41z2z4∣∣
∣
∣∣
Δ=xnynzn∣∣
∣
∣∣1x2x40y2−z2y4−z40z2−x2z4−x4∣∣
∣
∣∣(R2→R2−R3,R3→R3−R1)
Δ=xnynzn[(y2−z2)(z2+x2)(z2−x2) −(z2−x2)(y2+z2)(y2−z2)]Δ=xnynzn(y2−z2)(z2−x2)(x2+z2−y2−z2)Δ=xnynzn(x2−y2)(y2−z2)(z2−x2)Given that Δ=(1y2−1x2)(1z2−1y2)(1x2−1z2)
Δ=(x2−y2x2y2)(y2−z2y2z2)(z2−x2x2z2)Δ=x−4y−4z−4(x2−y2)(y2−z2)(z2−x2)
Hence n =-4