Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non-zero vectors such that no two of them are collinea

Question

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non-zero vectors such that no two of them are collinear and $(\vec{a} 	imes \vec{b}) 	imes \vec{c}=\frac{1}{3}|\vec{b}||\vec{c}| \vec{a}$. If $	heta$ is the angle between vectors $\vec{b}$ and $\vec{c}$, then a value of $\sin 	heta$ is\

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(c) : Expanding\ $(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{b} \cdot \vec{c}) \vec{a}=\frac{1}{3}|\vec{b}||\vec{c}| \vec{a}$\ $\Rightarrow \quad(\vec{a} \cdot \vec{c}) \vec{b}-\left\{(\vec{b} \cdot \vec{c})+\frac{1}{3}|\vec{b}||\vec{c}|ight\} \vec{a}=0$\ As $\vec{a}$ and $\vec{b}$ are non-collinear, the coefficients must vanish. Thus,\ $\vec{a} \cdot \vec{c}=0$ and $(\vec{b} \cdot \vec{c})=-\frac{1}{3}|\vec{b}||\vec{c}|$\ Again, $\cos 	heta=\frac{\vec{b} \cdot \vec{c}}{|\vec{b}||\vec{c}|} \Rightarrow \cos 	heta=-\frac{1}{3}$\ $\Rightarrow \quad \sin 	heta=\frac{2 \sqrt{2}}{3}$\