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(D) Given $|a+b+c|^2 = 1^2 = 1$. Since $c = \lambda(a 	imes b)$, $c$ is perpendicular to both $a$ and $b$, so $a \cdot c = 0$ and $b \cdot c = 0$.Exp

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$\frac{\pi}{2}$

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(D) Given $|a+b+c|^2 = 1^2 = 1$. Since $c = \lambda(a 	imes b)$, $c$ is perpendicular to both $a$ and $b$, so $a \cdot c = 0$ and $b \cdot c = 0$.Expanding $|a+b+c|^2 = |a|^2 + |b|^2 + |c|^2 + 2(a \cdot b + b \cdot c + c \cdot a) = 1$.Substituting the values: $(\frac{1}{\sqrt{3}})^2 + (\frac{1}{\sqrt{2}})^2 + (\frac{1}{\sqrt{6}})^2 + 2(a \cdot b + 0 + 0) = 1$.$\frac{1}{3} + \frac{1}{2} + \frac{1}{6} + 2(a \cdot b) = 1$.$\frac{2+3+1}{6} + 2(a \cdot b) = 1 \implies 1 + 2(a \cdot b) = 1 \implies a \cdot b = 0$.Since $a \cdot b = |a||b| \cos 	heta = 0$ and $|a|, |b| 
eq 0$, we have $\cos 	heta = 0$, so $	heta = \frac{\pi}{2}$.

If $a, b$ and $c$ are three vectors such that $|a + b + c| = 1$, $c = \lambda(a \times b)$, and $|a| = \frac{1}{\sqrt{3}}$, $|b| = \frac{1}{\sqrt{2}}$, $|c| = \frac{1}{\sqrt{6}}$, then the angle between $a$ and $b$ is

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