b→ and c→ are non-collinear If a→×(b→×c→)+(a→⋅b→)b→=(4−2x−sin⁡y)b→+x2−1c→ and (c→⋅c→)a→=c→ then

a→×(b→×c→)+(a→⋅b→)b→=(4−2x−sin⁡y)b→+x2−1c→ or (a→⋅c→)b→−(a→⋅b→)c→+(a→⋅b→)b→=(4−2x−sin⁡y)b→+x2−1c→now , (c→⋅c→)a→=c→. Therefore (c→⋅c→)(a→⋅c→)=(c→⋅c→) or a→⋅c→=1⇒1+a→⋅b→=4−2x−sin⁡y,x2−1=−(a→⋅b→)or 1=4−2x−sin⁡y+x2−1or sin⁡y=x2−2x+2=(x−1)2+1but sin⁡y≤1⇒x=1,sin⁡y=1⇒y=(4n+1)π2, n∈I

b→ and c→ are non-collinear If a→×(b→×c→)+(a→⋅b→)b→=(4−2x−sin⁡y)b→+x2−1c→ and (c→⋅c→)a→=c→ then

a→×(b→×c→)+(a→⋅b→)b→=(4−2x−sin⁡y)b→+x2−1c→ or (a→⋅c→)b→−(a→⋅b→)c→+(a→⋅b→)b→=(4−2x−sin⁡y)b→+x2−1c→now , (c→⋅c→)a→=c→. Therefore (c→⋅c→)(a→⋅c→)=(c→⋅c→) or a→⋅c→=1⇒1+a→⋅b→=4−2x−sin⁡y,x2−1=−(a→⋅b→)or 1=4−2x−sin⁡y+x2−1or sin⁡y=x2−2x+2=(x−1)2+1but sin⁡y≤1⇒x=1,sin⁡y=1⇒y=(4n+1)π2, n∈I

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$\vec{b} and \vec{c}$ are non-collinear If $\vec{a} \times (\vec{b} \times \vec{c}) + (\vec{a} \cdot \vec{b}) \vec{b} = (4 - 2 x - \sin y) \vec{b} + (x^{2} - 1) \vec{c}$ and $(\vec{c} \cdot \vec{c}) \vec{a} = \vec{c}$ then

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x=1

x=-1

$y = (4 n + 1) \frac{π}{2}, n \in I$

$y = (2 n + 1) \frac{π}{2}, n \in I$

answer is A, C.

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Detailed Solution

$\begin{array}{l} \vec{a} \times (\vec{b} \times \vec{c}) + (\vec{a} \cdot \vec{b}) \vec{b} \\ = (4 - 2 x - \sin y) \vec{b} + (x^{2} - 1) \vec{c} \\ or (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} + (\vec{a} \cdot \vec{b}) \vec{b} \\ = (4 - 2 x - \sin y) \vec{b} + (x^{2} - 1) \vec{c} \end{array}$

now , $(\vec{c} \cdot \vec{c}) \vec{a} = \vec{c} . Therefore$

$\begin{array}{l} (\vec{c} \cdot \vec{c}) (\vec{a} \cdot \vec{c}) = (\vec{c} \cdot \vec{c}) or \vec{a} \cdot \vec{c} = 1 \\ \Rightarrow 1 + \vec{a} \cdot \vec{b} = 4 - 2 x - \sin y, x^{2} - 1 = - (\vec{a} \cdot \vec{b}) \\ or 1 = 4 - 2 x - \sin y + x^{2} - 1 \\ or \sin y = x^{2} - 2 x + 2 = (x - 1)^{2} + 1 \\ but \sin y \leq 1 \Rightarrow x = 1, \sin y = 1 \\ \Rightarrow y = (4 n + 1) \frac{π}{2}, n \in I \end{array}$