Plz solve

​Q48. Show that the vectors $\overrightarrow{\mathrm{a}} ,\overrightarrow{\mathrm{b}} \mathrm{and} \overrightarrow{\mathrm{c}}$ are coplanar, if $\overrightarrow{\mathrm{a}} + \overrightarrow{\mathrm{b}} ,\overrightarrow{\mathrm{b}} +\overrightarrow{\mathrm{c}} \mathrm{and} \overrightarrow{\mathrm{c}} +\overrightarrow{\mathrm{a}}$ are coplanar. 
                          Or
  Prove that, for any three vectors ​$\overrightarrow{\mathrm{a}} ,\overrightarrow{\mathrm{b}} \mathrm{and} \overrightarrow{\mathrm{c}}$,​$\left[\overrightarrow{\mathrm{a}} + \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{b}} +\overrightarrow{\mathrm{c}} \overrightarrow{\mathrm{c}} +\overrightarrow{\mathrm{a}}\right] =2\left[\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}\right]$.

For the 48 th question, both the first and OR question are related to each other. 
1. You have to add the first statement [a b c] = 0. then proceed as told. To prove this we write box product of all the vectors to be proven coplanar i.e. [(a+b) (b+c) (c+a)]. open this completely and write 0 whenever [a b c]  or box product with 2 same vectors appear. final answer is 0. therefore (a+b), (b+c), (c+a) is coplanar.

2. start by ​ [(a+b) (b+c) (c+a)]. open it but only substitute 0 for box product with 2 same vectors. your final result would be 2[a b c].