Let $\overrightarrow{a}=\widehat{\mathrm{i}}+\widehat{\mathrm{j}}+\sqrt{2}\widehat{\mathrm{k}},$ $\overrightarrow{b}=b_1\widehat{\mathrm{i}}+b_2\widehat{\mathrm{j}}+\sqrt{2}\widehat{\mathrm{k}}$ and $\overrightarrow{c}=5\widehat{\mathrm{i}}+\widehat{\mathrm{j}}+\sqrt{2}\widehat{\mathrm{k}}$ be three vectors such that the projection vector of $\overrightarrow{b}$ on $\overrightarrow{a}$ is $\left|\overrightarrow{a}\right|$ . If $\overrightarrow{a}+\overrightarrow{b}$ is perpendicular to $\overrightarrow{c}$ , then $\left|\overrightarrow{b}\right|$ is equal to:

• A

  $\sqrt{22}$
• B

  $\sqrt{32}$
• Correct answer
  C

  $6$
• D

  $4$

Hence, the correct answer is option: 'C'.