If $\vec{a}$, $\vec{b}$, $\vec{c}$ are unit vectors such that $\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} = 0$, and the angle between $\vec{b}$ and $\vec{c}$ is $\frac{\pi}{6}$, then

Given $\vec{a}$, $\vec{b}$, $\vec{c}$ are unit vectors with $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$.

From $\vec{a} \cdot \vec{b} = 0$, the vector $\vec{a}$ is perpendicular to $\vec{b}$.

From $\vec{a} \cdot \vec{c} = 0$, the vector $\vec{a}$ is perpendicular to $\vec{c}$.

The angle between $\vec{b}$ and $\vec{c}$ is $\frac{\pi}{6}$.

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The cross product $\vec{b} \times \vec{c}$ is perpendicular to both $\vec{b}$ and $\vec{c}$.

Since $\vec{a}$ is also perpendicular to both $\vec{b}$ and $\vec{c}$, the vectors $\vec{a}$ and $\vec{b} \times \vec{c}$ are parallel.

Therefore:

$\vec{a} = k(\vec{b} \times \vec{c})$

where $k$ is a scalar.

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The magnitude of the cross product is:

$|\vec{b} \times \vec{c}| = |\vec{b}| \cdot |\vec{c}| \cdot \sin\theta$

$|\vec{b} \times \vec{c}| = 1 \cdot 1 \cdot \sin\frac{\pi}{6}$

$|\vec{b} \times \vec{c}| = \frac{1}{2}$

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Taking magnitudes in $\vec{a} = k(\vec{b} \times \vec{c})$:

$|\vec{a}| = |k| \cdot |\vec{b} \times \vec{c}|$

$1 = |k| \cdot \frac{1}{2}$

$|k| = 2$

Therefore $k = \pm 2$.

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$\vec{a} = \pm 2(\vec{b} \times \vec{c})$