$({\cos }^{-1}l+{\cos }^{-1}m+{\cos }^{-1}n)$ is equal to

The correct option is B ${180}^o$
For positive $l,m$ and $n$, if the planes $x=ny+mx,y=lz+nx,z=mx+ly$ intersect in a straight line.

$\Rightarrow$ given system has to have infinitely many solutions, non-trivial solutions.

$\therefore \left|\begin{array}{ccc}-1& n& m\\ n& -1& l\\ m& l& -1\end{array}\right|=0$

$\Rightarrow l^2+m^2+n^2+2lmn=1$

Possible values of $l,m,n$ are $l=1,m=1,n=-1$

$({\cos }^{-1}l+{\cos }^{-1}m+{\cos }^{-1}n)=0+0+\pi =\pi$

Hence, option C is correct.