1. **State the problem:** Given three non-zero vectors $a$, $b$, and $c$, where $c$ is a unit vector perpendicular to both $a$ and $b$, and the angle between $a$ and $b$ is $\frac{\pi}{6}$, show that $$[a,b,c]^2 = \frac{1}{4} |a|^2 |b|^2.$$ 2. **Recall the scalar triple product definition:** The scalar triple product $[a,b,c]$ is defined as $$[a,b,c] = a \cdot (b \times c).$$ 3. **Use the property of $c$ being perpendicular to $a$ and $b$:** Since $c$ is perpendicular to both $a$ and $b$, $c$ is perpendicular to the plane containing $a$ and $b$. Also, $c$ is a unit vector, so $|c|=1$. 4. **Express $b \times c$:** The magnitude of $b \times c$ is $$|b \times c| = |b||c|\sin\theta = |b| \sin\theta,$$ where $\theta$ is the angle between $b$ and $c$. Since $c$ is perpendicular to $b$, $\theta = \frac{\pi}{2}$, so $$|b \times c| = |b| \cdot 1 = |b|.$$ 5. **Calculate $[a,b,c]$:** Since $[a,b,c] = a \cdot (b \times c)$, and $b \times c$ is perpendicular to $c$ and lies in the plane of $a$ and $b$, the vector $b \times c$ is parallel to $a$ (or anti-parallel). Thus, $$[a,b,c] = |a||b \times c| \cos\phi,$$ where $\phi$ is the angle between $a$ and $b \times c$. 6. **Determine the angle $\phi$:** Because $c$ is perpendicular to $a$ and $b$, $b \times c$ is parallel to $a$. Therefore, $\phi = 0$ and $$\cos\phi = 1.$$ 7. **Substitute values:** $$[a,b,c] = |a||b| \cdot 1 = |a||b|.$$ 8. **Check the problem statement:** The problem states the angle between $a$ and $b$ is $\frac{\pi}{6}$, so we need to incorporate this. 9. **Use the vector identity for scalar triple product:** Recall that $$[a,b,c] = (a \times b) \cdot c.$$ Since $c$ is perpendicular to $a$ and $b$, and a unit vector, it is in the direction of $a \times b$ normalized: $$c = \frac{a \times b}{|a \times b|}.$$ 10. **Calculate $[a,b,c]$ using this:** $$[a,b,c] = (a \times b) \cdot c = (a \times b) \cdot \frac{a \times b}{|a \times b|} = |a \times b|.$$ 11. **Calculate $|a \times b|$:** $$|a \times b| = |a||b| \sin \theta = |a||b| \sin \frac{\pi}{6} = |a||b| \cdot \frac{1}{2} = \frac{1}{2} |a||b|.$$ 12. **Square both sides:** $$[a,b,c]^2 = \left(\frac{1}{2} |a||b|\right)^2 = \frac{1}{4} |a|^2 |b|^2.$$ **Final answer:** $$[a,b,c]^2 = \frac{1}{4} |a|^2 |b|^2.$$