Let $$\left| {\overrightarrow a } \right| = 1,\,\left| {\overrightarrow b } \right| = \sqrt 2 ,\,\left| {\overrightarrow c } \right| = \sqrt 3 ,$$       and $$\,\overrightarrow a \bot \left( {\overrightarrow b + \overrightarrow c } \right),\,\overrightarrow b \bot \left( {\overrightarrow c + \overrightarrow a } \right)$$       and $$\overrightarrow c \bot \left( {\overrightarrow a + \overrightarrow b } \right).$$    Then $$\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|$$   is :

Solution :
$$\overrightarrow a .\left( {\overrightarrow b + \overrightarrow c } \right) = 0{\text{ or }}\overrightarrow a .\overrightarrow b + \overrightarrow a .\overrightarrow c = 0$$
Similarly, we get $$\overrightarrow b .\overrightarrow c + \overrightarrow b .\overrightarrow a = 0$$     and $$\overrightarrow c .\overrightarrow a + \overrightarrow c .\overrightarrow b = 0$$
Adding these, $$\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c
+ \overrightarrow c .\overrightarrow a = 0$$
$$\eqalign{ & {\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|^2} = {\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right)^2} \cr & = {\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2} + {\left| {\overrightarrow c } \right|^2} + 2\left( {\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a } \right) \cr & = 1 + 2 + 3 + 0 \cr & = 6 \cr & \therefore \,\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right| = \sqrt 6 \cr} $$