Given $(a \times b) \times (c \times d) = \frac{1}{6}i - \frac{1}{3}j + \frac{1}{3}k$
$ \Rightarrow [(a \times b)\,.\,d]\,c - [(a \times b)\,.\,c]d = \frac{1}{6}i - \frac{1}{3}j + \frac{1}{3}k$
$ \Rightarrow [(|a||b|\sin 30^\circ )\,\hat n\,.\,d]\,c - 0 = \frac{1}{6}i - \frac{1}{3}j + \frac{1}{3}k$
$ \Rightarrow \left[ {(1)(1)\left( {\frac{1}{2}} \right)} \right][|\hat n||d|\cos \theta ]\,c = \frac{1}{6}i - \frac{1}{3}j + \frac{1}{3}k$
$\Rightarrow \frac{1}{2}\cos \theta \,c=\frac{1}{6}i-\frac{1}{3}j+\frac{1}{3}k$
Where $\hat n$ and $d$ are unit perpendicular vector and angle between $\hat n$ and $d$ may be $ 0 $ or $\pi $.
When $\theta = 0^\circ ,$ $c = \frac{1}{3}[i - 2j + 2k]$
When $\theta = \pi ,$ $c = \frac{1}{3}[ - i + 2j - 2k]$.
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