Answer:
Option A
Explanation:
Given,
(a x b)x c= $\frac{1}{3}|b||c|a.$
$\Rightarrow$ -c x (a x b) = $\frac{1}{3}|b||c|a.$
$\Rightarrow$ $-(c.b).a+(c.a)b= \frac{1}{3}|b||c|a$
$\left[ \frac{1}{3}|b||c|+(c.b)\right]a=(c.a)b$
Since a and b are not collinear
$c.b +\frac{1}{3}|b||c|=0 $ and c.a=0
$\Rightarrow$ $|c||b|\cos \theta +\frac{1}{3}|b||c|=0 $
$\Rightarrow$ $|b||c|\left(\cos \theta +\frac{1}{3}\right)=0 $
$\Rightarrow$ $\cos \theta +\frac{1}{3}=0, (\therefore |b|\neq0,|c|\neq0)$
$\Rightarrow$ $\cos \theta =-\frac{1}{3}$
$\Rightarrow$ $\sin \theta =\frac{\sqrt{8}}{3}=\frac{2\sqrt{2}}{3}$