The correct option is D 5π6
|→a|=1,|→b|=4,→a⋅→b=2,→c=(2→a×→b)−3→b
⇒→c+3→b=2→a×→b∵→a⋅→b=2⇒cosθ=24=12⇒θ=π3⇒|→c+3→b|2=|2→a×→b|2
Taking →b⋅→c=|→b||→c|cosα
⇒|→c|2+144+6→b⋅→c=48⇒|→c|2+96+6(→b⋅→c)=0
→c=(2→a×→b)−3→b
⇒→b⋅→c=0−3×16=−48
⇒|→c|2+96−6×48=0⇒|→c|2=192
Taking →b⋅→c=|→b||→c|cosα
⇒192+96+6|→b|⋅|→c|cosα=0⇒cosα=−2886×4×8√3=−32√3=−√32⇒α=5π6