Let $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\sqrt{2} \hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}={\mathrm{b}}_1\hat{\mathrm{i}}+{\mathrm{b}}_2\hat{\mathrm{j}}+\sqrt{2} \hat{\mathrm{k}}$ and $\overrightarrow{c}=5\hat{\mathrm{i}}+\hat{\mathrm{j}}+\sqrt{2} \hat{\mathrm{k}}$ be three vectors such that the projection vector of $\overrightarrow{\mathrm{b}}$ on $\overrightarrow{\mathrm{a}}$ is  $\overrightarrow{\mathrm{a}}$. If $\overrightarrow{\mathrm{a}}$+$\overrightarrow{\mathrm{b}}$ is perpendicular to $\overrightarrow{\mathrm{c}}$ then $\left|\overrightarrow{\mathrm{b}}\right|$ is equal to 

$$\begin{gathered} \left(\frac{\overline{\mathrm{b}}.\overline{\mathrm{a}}}{{\left|\overline{\mathrm{a}}\right|}^2}\right)\overline{\mathrm{a}}=\overline{\mathrm{a}} \Rightarrow \frac{\overline{\mathrm{b}}.\overline{\mathrm{a}}}{{\left|\overline{\mathrm{a}}\right|}^2}=1 \\ {\mathrm{b}}_1+{\mathrm{b}}_2+2=1+1+2 \\ {\mathrm{b}}_1+{\mathrm{b}}_2=2 -\left(1\right) \\ (\overline{\mathrm{a}}+\overline{\mathrm{b}}).\overline{\mathrm{c}}=0 \Rightarrow \overline{\mathrm{a}}.\overline{\mathrm{c}}+\overline{\mathrm{b}}.\overline{\mathrm{c}}=0 \\ 5+1+2+5{\mathrm{b}}_1{\mathrm{b}}_2+2=0 \\ 5{\mathrm{b}}_1+2-{\mathrm{b}}_1+10=0 \Rightarrow 4{\mathrm{b}}_1=-12 \\ {\mathrm{b}}_1=-3 \\ {\mathrm{b}}_2=5 \\ \left|\overline{\mathrm{b}}\right|=\sqrt{9+25+2}=\sqrt{36}=0 \end{gathered}$$