For a complex number $Z=e^{i \theta}$ it is given that
$\begin{aligned} & \frac{3 \cos 3 \theta+2 \cos 2 \theta+5 \cos 5 \theta}{3 \sin 3 \theta+2 \sin 2 \theta+5 \sin 5 \theta}=i \frac{\sum_{r=0}^{10} a_r Z^r}{\sum_{r=0}^{10} b_r Z^r} \\ & \Rightarrow \frac{\sum_{r=0}^{10} a_r z^r+\sum_{r=0}^{10} b_r z^r}{\sum_{r=0}^{10} a_r z^r-\sum_{r=0}^{10} b_r z^r}=\frac{3 e^{i 3 \theta}+2 e^{i 2 \theta}+5 e^{i 5 \theta}}{3 e^{-i 3 \theta}+2 e^{-i 2 \theta}+5 e^{-i 5 \theta}} \\ & \sum_{r=0}^{10} z^r\left(a_r+b_r\right) \\ & \sum_{r=0}^{10} z^r\left(a_r-b_r\right) \quad=\frac{3 e^{i 3 \theta}+2 e^{i 2 \theta}+5 e^{i 5 \theta}}{3 e^{-i 3 \theta}+2 e^{-i 2 \theta}+5 e^{-i 5 \theta}} \\ & \Rightarrow a_0+b_0=a_1+b_1=a_4+b_4=a_6+b_6 \\ & =a_7+b_7=a_8+b_8 \\ & =a_9+b_9=a_{10}+b_{10}=0 \\ & \text { and } a_2+b_2=2, a_3+b_3=3, a_5+b_5=5\end{aligned}$
$\therefore \quad \frac{\sum_{r=0}^{10}\left(a_r+b_r\right)}{10}=\frac{2+3+5}{10}=1$