We have.

$\tan \left(\theta+\frac{\pi}{4}\right)=3 \tan 3 \theta$

$\Rightarrow \frac{1+\tan \theta}{1-\tan \theta}=3 \cdot \frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}$

$\Rightarrow \frac{1+t}{1-t}=3\left(\frac{3 t-t^{3}}{1-3 t^{2}}\right) \quad[\text { putting } t=\tan \theta]$

$\Rightarrow 3 t^{4}-6 t^{2}+8 t-1=0$

$\mathrm{So}, \mathrm{t}_{1}+\mathrm{t}_{2}+\mathrm{t}_{3}+\mathrm{t}_{4}=\frac{0}{3}=0$

$\therefore $ $\tan \alpha+\tan \beta+\tan \gamma+\tan \delta=0$

Hence, $( 4)$ is the correct answer.