Let $$\vec{a}, \vec{b}$$ and $$\vec{c}$$ be non-zero vectors such that $$(\vec{a} \times \vec{b}) \times \vec{c} = \frac{1}{3}|\vec{b}||\vec{c}|\vec{a}$$. If $$\theta$$ is the acute angle between the vectors $$\vec{b}$$ and $$\vec{c}$$, then $$\sin\theta$$ equals

Using the vector triple product identity,

$$(\vec a\times\vec b)\times\vec c=\vec b(\vec a\cdot\vec c)-\vec a(\vec b\cdot\vec c)$$

Given,

$$\vec b(\vec a\cdot\vec c)-\vec a(\vec b\cdot\vec c)=\frac13|\vec b||\vec c|\vec a$$

Therefore,

$$\vec b(\vec a\cdot\vec c)=\left(\vec b\cdot\vec c+\frac13|\vec b||\vec c|\right)\vec a$$

Since $$\vec a$$ and $$\vec b$$ are non-zero vectors and the RHS is a scalar multiple of $$\vec a$$ while the LHS is a scalar multiple of $$\vec b$$, we must have

$$\vec a\cdot\vec c=0$$

and

$$\vec b\cdot\vec c+\frac13|\vec b||\vec c|=0$$

Hence,

$$\vec b\cdot\vec c=-\frac13|\vec b||\vec c|$$

Therefore,

$$\cos\theta=\frac{|\vec b\cdot\vec c|}{|\vec b||\vec c|}=\frac13$$

Since $$\theta$$ is acute,

$$\cos\theta=\frac13$$

Thus,

$$\sin\theta=\sqrt{1-\cos^2\theta}$$

$$=\sqrt{1-\frac19}$$

$$=\sqrt{\frac89}$$

$$=\frac{2\sqrt2}{3}$$

Hence,

$$\boxed{\frac{2\sqrt2}{3}}$$