If cos⁡α+2cos⁡β+3cos⁡γ=sin⁡α+2sin⁡β+3sin⁡γ=0 then the value of sin 3α+8sin⁡3β+27sin⁡3γ is

Let a=cos⁡α+isin⁡αb=cos⁡β+isin⁡βc=cos⁡γ+isin⁡γThen, a+2b+3c=(cos⁡α+2cos⁡β+3cos⁡γ)                                        +i(sin⁡α+2sin⁡β+3sin⁡γ)=0⇒a3+8b3+27c3=18abc⇒cos⁡3α+8cos⁡3β+27cos⁡3γ=18cos⁡(α+β+γ)and sin⁡3α+8sin⁡3β+27sin⁡3γ=18sin⁡(α+β+γ)