Question:

Step 1: Calculate $\lambda$. Given that $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ are unit vectors, we have $|\mathbf{a}|=|\mathbf{b}|=|\mathbf{c}|=1$. Given $\mathbf{a}+\mathbf{b}+\mathbf{c}=\mathbf{0}$. Squaring both sides of the equation: $$|\mathbf{a}+\mathbf{b}+\mathbf{c}|^2 = |\mathbf{0}|^2$$ $$(\mathbf{a}+\mathbf{b}+\mathbf{c}) \cdot (\mathbf{a}+\mathbf{b}+\mathbf{c}) = 0$$ $$|\mathbf{a}|^2 + |\mathbf{b}|^2 + |\mathbf{c}|^2 + 2(\mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c} + \mathbf{c} \cdot \mathbf{a}) = 0$$ Substituting the magnitudes and the definition of $\lambda$: $$1^2 + 1^2 + 1^2 + 2\lambda = 0$$ $$3 + 2\lambda = 0$$ $$\lambda = -\frac{3}{2}$$ Step 2: Analyze the geometric configuration of the vectors. Since $\mathbf{a}+\mathbf{b}+\mathbf{c}=\mathbf{0}$ and $|\mathbf{a}|=|\mathbf{b}|=|\mathbf{c}|=1$, these three vectors must form an equilateral triangle when placed head-to-tail. This implies that the angle between any pair of these vectors is $120^\circ$. For example, consider $\mathbf{a}+\mathbf{b}=-\mathbf{c}$. Squaring both sides: $$|\mathbf{a}+\mathbf{b}|^2 = |-\mathbf{c}|^2$$ $$|\mathbf{a}|^2 + |\mathbf{b}|^2 + 2(\mathbf{a} \cdot \mathbf{b}) = |\mathbf{c}|^2$$ $$1^2 + 1^2 + 2(\mathbf{a} \cdot \mathbf{b}) = 1^2$$ $$2 + 2(\mathbf{a} \cdot \mathbf{b}) = 1$$ $$2(\mathbf{a} \cdot \mathbf{b}) = -1$$ $$\mathbf{a} \cdot \mathbf{b} = -\frac{1}{2}$$ Since $|\mathbf{a}|=|\mathbf{b}|=1$, we have $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta = \cos\theta$. Thus, $\cos\theta = -\frac{1}{2}$, which implies $\theta = 120^\circ$. The same applies to $\mathbf{b} \cdot \mathbf{c}$ and $\mathbf{c} \cdot \mathbf{a}$. Step 3: Calculate $\mu$. The magnitude of the cross product of two unit vectors with an angle $\theta$ between them is given by $|\mathbf{u} \times \mathbf{v}| = |\mathbf{u}||\mathbf{v}|\sin\theta$. For $\mathbf{a} \times \mathbf{b}$, the angle is $120^\circ$: $$|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin(120^\circ) = (1)(1)\frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$$ Similarly, for $\mathbf{b} \times \mathbf{c}$ and $\mathbf{c} \times \mathbf{a}$: $$|\mathbf{b} \times \mathbf{c}| = \frac{\sqrt{3}}{2}$$ $$|\mathbf{c} \times \mathbf{a}| = \frac{\sqrt{3}}{2}$$ Now, calculate $\mu$: $$\mu = |\mathbf{a} \times \mathbf{b}| + |\mathbf{b} \times \mathbf{c}| + |\mathbf{c} \times \mathbf{a}| = \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}$$ Step 4: Conclusion. The calculated values are $\lambda = -\frac{3}{2}$ and $\mu = \frac{3\sqrt{3}}{2}$. Such a configuration of three unit vectors lying in a plane, with $120^\circ$ between each pair, is geometrically possible. Therefore, the values $(\lambda, \mu) = \left(-\frac{3}{2}, \frac{3\sqrt{3}}{2}\right)$ are valid.