(B) Given the equation: $\tan \theta + \tan 2 \theta + \tan 3 \theta = 0$. Since $3 \theta = \theta + 2 \theta$, we have $\tan 3 \theta = \tan (\theta

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(B) Given the equation: $\tan \theta + \tan 2 \theta + \tan 3 \theta = 0$. Since $3 \theta = \theta + 2 \theta$, we have $\tan 3 \theta = \tan (\theta + 2 \theta) = \frac{\tan \theta + \tan 2 \theta}{1 - \tan \theta \cdot \tan 2 \theta}$. Substituting $\tan \theta + \tan 2 \theta = -\tan 3 \theta$ and $\tan \theta \cdot \tan 2 \theta = k$ into the formula: $\tan 3 \theta = \frac{-\tan 3 \theta}{1 - k}$. Since $\tan 3 \theta \neq 0$, we can divide both sides by $\tan 3 \theta$: $1 = \frac{-1}{1 - k}$. $1 - k = -1$. $k = 2$.