If cos−1⁡α+cos−1⁡β+cos−1⁡γ=3π , then α(β+γ)+β(γ+α)+γ(α+β) equals

We have, cos−1⁡α+cos−1⁡β+cos−1⁡γ=3π We know that, 0≤cos−1⁡x≤π⇒cos−1⁡α+cos−1⁡β+cos−1⁡γ=3π is possible if and only if cos−1⁡α=cos−1⁡β=cos−1⁡γ=π⇒α=β=γ=−1∴α(β+γ)+β(γ+α)+γ(α+β)=−1(−1−1)−1(−1−1)−1(−1−1)=6