TS EAMCET 11-Sep-2020 Shift 1 Solved Paper

For non-collinear vectors b and c,|c|≠0, it is given that, (c⋅c)a=c⇒a⋅c=1 ...(i)
and a×(b×c)+(a⋅b)b
=(4−2β−sin‌α)b+(β2−1)c
So, a⋅c+a⋅b=4−2β−sin‌α....(ii)
and ‌‌−a⋅b=β2−1.....(ii)
From Eqs. (i), (ii) and (iii), we get
1=4−2β−sin‌α+β2−1
⇒‌‌(β−1)2−sin‌α=−1
So, if β=1, then sin‌α=1⇒α=2nπ+‌

π

2

,n∈Z
and other options does not satisfy.