Since, the vector a=(1,3,sin‌2‌α) makes an obtuse angle with the Z - axis. Therefore, its z - component is negative. i.e. sin‌2‌α<0 ∴ −1≤sin‌2‌α<0....(i) Since, b and c are orthogonal. ∴ b.c=0 ⇒ tan2α−tan‌α−6=0 ⇒ (tan‌α−3)(tan‌α+2)=0 ⇒ tan‌α=3,−2 ∴ tan‌α=3 Then, sin‌2‌α=
2‌tan‌α
1+tan2α
=
6
10
>0, which is a contradiction to ......(i) ∴ tan‌α=3 is not possible. Thus, tan‌α=−2 and for this value of tan‌α, we get tan‌2‌α=
2‌tan‌α
1−tan2α
=
4
3
Since, sin‌2‌α<0‌and‌tan‌2‌α>0 Therefore, 2α is in third quadrant. Also, √sin‌
α
2
is meaningful, if 0<sin‌
α
2
<1 when these conditions are satisfied, α is given by α=(4n+1)π−tan−12.