Given the three vectors are coplanar.

Hence, $\left|\begin{array}{ccc}\mu & 1& 1\\ 1& \mu & 1\\ 1& 1& \mu \end{array}\right|=0$

$\Rightarrow \mu \left({\mu }^2-1\right)-1\left(\mu -1\right)+1\left(1-\mu \right)=0$

$\Rightarrow \mu \left(\mu -1\right)\left(\mu +1\right)-2\left(\mu -1\right)=0 \left[\because a^2-b^2=\left(a-b\right)\left(a+b\right)\right]$

$\Rightarrow \left(\mu -1\right)\left(\mu \left(\mu +1\right)-2\right)=0$

$\Rightarrow \left(\mu -1\right)\left({\mu }^2+\mu +2\right)=0$

$\Rightarrow \left(\mu -1\right)\left(\mu -1\right)\left(\mu +2\right)=0$

$\Rightarrow \mu =1, 1, -2$

Therefore, the sum of the distinct real values of $\mu =-2+1=-1$.