Three planes intersect in a line ∴D=D1=D2=D3=0
Here D1=D2=D3=0,∴D=0 Δ=0 ∣∣−1sinαsinβsinα−1sinθsinβsinθ−1∣∣=0 −1(1−sin2θ)−sinα(−sinα−sinθsinβ) +sinβ(sinαsinθ+sinβ)=0 sin2θ+sin2α+sin2β+2sinαsinβsinθ=1
Now D.R's of line ∣∣i−1sinαjsinα−1ksinβsinθ∣∣ =i(sinαsinθ+sinβ)+j(sinαsinβ+sinθ)+k(1−sin2α)
This vector is parallel to line. ∴cosθsinαsinθ+sinβ=cosβsinαsinβ+sinθ=cosα1−sin2θ sinαsinθ+sinβ=cosθcosα cos(θ+α)=cos(2π−β) ∴α+β+θ=2nπ+2π