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(D) Since $a, b, c$ are coplanar,their scalar triple product $[a, b, c] = 0$.Given the vector triple product identity $(a 	imes b) 	imes (c 	imes d

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$A$ and $C$

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(D) Since $a, b, c$ are coplanar,their scalar triple product $[a, b, c] = 0$.Given the vector triple product identity $(a 	imes b) 	imes (c 	imes d) = [(a 	imes b) \cdot d]c - [(a 	imes b) \cdot c]d$.Since $c$ is coplanar with $a$ and $b$,$[a, b, c] = (a 	imes b) \cdot c = 0$,so the second term vanishes.Thus,$[(a 	imes b) \cdot d]c = \frac{1}{6}i - \frac{1}{3}j + \frac{1}{3}k$.Let $n$ be the unit vector perpendicular to $a$ and $b$. Then $a 	imes b = |a||b| \sin(30^\circ) n = (1)(1)(\frac{1}{2})n = \frac{1}{2}n$.Substituting this,we get $\frac{1}{2}(n \cdot d)c = \frac{1}{6}i - \frac{1}{3}j + \frac{1}{3}k$.Since $n$ and $d$ are both unit vectors perpendicular to the plane of $a, b, c$,$n \cdot d = \pm 1$.If $n \cdot d = 1$,then $\frac{1}{2}c = \frac{1}{6}i - \frac{1}{3}j + \frac{1}{3}k \Rightarrow c = \frac{1}{3}i - \frac{2}{3}j + \frac{2}{3}k = \frac{i - 2j + 2k}{3}$.If $n \cdot d = -1$,then $-\frac{1}{2}c = \frac{1}{6}i - \frac{1}{3}j + \frac{1}{3}k \Rightarrow c = -\frac{1}{3}i + \frac{2}{3}j - \frac{2}{3}k = \frac{-i + 2j - 2k}{3}$.Both options $A$ and $C$ are valid,so the correct option is $D$.

Unit vectors $a, b$ and $c$ are coplanar. $A$ unit vector $d$ is perpendicular to them. If $(a \times b) \times (c \times d) = \frac{1}{6}i - \frac{1}{3}j + \frac{1}{3}k$ and the angle between $a$ and $b$ is $30^\circ$,then $c$ is:

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