If $x+y=z$, then ${\cos }^{2}x+{\cos }^{2}y+{\cos }^{2}z-2\cos x\cos y\cos z$ is equal to 

The given expression can be written as

$\begin{array}{l}{\cos }^{2}x+{\cos }^{2}y+{\cos }^{2}z-\cos z[\cos (x+y)+\cos (x-y\left)\right]\\ ={\cos }^{2}x+{\cos }^{2}y+{\cos }^{2}z-{\cos }^{2}z-\cos (x+y)\cos (x-y)\end{array}$

$\begin{array}{l}={\cos }^{2}x+{\cos }^{2}y-(1/2)[\cos 2x+\cos 2y]\\ =(1/2)\left[2{\cos }^{2}x+2{\cos }^{2}y-\cos 2x-\cos 2y\right]\\ =(1/2)\left[2{\cos }^{2}x+2{\cos }^{2}y-2{\cos }^{2}x+1-2{\cos }^{2}y+1\right]=1\end{array}$