If →a,→b,→c are three non-zero vectors, no two of which are collinear, →a+→b is collinear with →c and →b+3→c is collinear with →a, then |→a+2→b+6→c| will be equal to
zero
→a+2→b=λ→c...(i)→b+3→c=μ→a,......(ii)
where no two of →a, →b and →c are collinear vectors
Eliminating →b from above relations,
⇒→a−6→c=λ→c−2μ→a⇒→a(1+2μ)=(λ+6)→c
Since →a, →b are non-collinear and non-zero,
1+2μ=0, λ+6=0⇒μ=−12, λ=−6
→a+2→b=−6→c⇒→a+2→b+6→c=→0