Since $\left(\overrightarrow{a}+\overrightarrow{b}\right)$ is perpendicular to $\overrightarrow{c}$, hence

$\left(\overrightarrow{a}+\overrightarrow{b}\right)·\overrightarrow{c}=0$

$\Rightarrow \overrightarrow{a}·\overrightarrow{c}+\overrightarrow{b}·\overrightarrow{c}=0$

$\Rightarrow 5+1+2+5b_1+b_2^{ }+2=0$

$\Rightarrow 5b_1+b_2=-10 ...\left(1\right)$ 

Given,

$\frac{\overrightarrow{b}·\overrightarrow{a}}{\left|\overrightarrow{a}\right|}=\left|\overrightarrow{a}\right|$

$\left\{\because \overrightarrow{p}=r\hat{\mathrm{i}}+s\hat{\mathrm{j}}+t\hat{\mathrm{k}}\Rightarrow \left|\overrightarrow{p}\right|=\sqrt{r^2+s^2+t^2}\right\}$

$\Rightarrow b_1+b_2+2=4$

$\Rightarrow b_1+b_2=2 ...\left(2\right)$ 

Solving $\left(1\right)$ and $\left(2\right),$ $b_1=-3, b_2=5$

$\therefore \left|\overrightarrow{b}\right|=\sqrt{9+25+2}=6$