Coordinate proof that the sum of a triangle's angles is $180^\circ$?

By the assumption of $P,Q,R$, one has that $x,y,$ and $z=-x-y$ are nonzero vectors in ${\mathbb R}^n$. Then one checks that $$|z|^2=(-x-y)\cdot(-x-y)=|x|^2+|y|^2+2x\cdot y,\quad (1)$$ and $$|x|^2|y|^2-(x\cdot y)^2=|y|^2|z|^2-(y\cdot z)^2=|z|^2|x|^2-(z\cdot x)^2.\quad (2)$$ Let $A,B,C$ be the angles you defined, namely $$A=\cos^{-1}\left(-\frac{x\cdot y}{|x||y|}\right),B=\cos^{-1}\left(-\frac{y\cdot z}{|y||z|}\right),C=\cos^{-1}\left(-\frac{z\cdot x}{|z||x|}\right).\quad (3)$$ By assumption $0<A+B+C<3\pi$. It follows that $$A+B+C=\pi$$ $$\Leftrightarrow \cos(A+B+C)=-1.\quad (4)$$ Therefore it suffices to show (4), but by trigonometric identities, one has $$\cos(A+B+C)=\cos(A)\cos(B)\cos(C)-\sum_{\rm cyc.}\sin(A)\sin(B)\cos(C)$$
$$=\left(\frac{-x\cdot y}{|x||y|}\right)\left(\frac{-y\cdot z}{|y||z|}\right)\left(\frac{-z\cdot x}{|z||x|}\right)-\sum_{\rm cyc.}\sqrt{1-\left(\frac{x\cdot y}{|x||y|}\right)^2}\sqrt{1-\left(\frac{y\cdot z}{|y||z|}\right)^2}\left(\frac{-z\cdot x}{|z||x|}\right),$$ which can be rewritten as (using (2) and factoring out the common factor) $$\frac{-(x\cdot y)(y\cdot z)(z\cdot x)}{|x|^2|y|^2|z|^2}-\frac{|x|^2|y|^2-(x\cdot y)^2}{|x|^2|y|^2|z|^2}(-z\cdot x-x\cdot y-y\cdot z)$$ $$=\frac{(-x\cdot y)(|y|^2+x\cdot y)(|x|^2+x\cdot y)}{|x|^2|y|^2|z|^2}-\frac{|x|^2|y|^2-(x \cdot y)^2}{|x|^2|y|^2|z|^2}(|z|^2-x\cdot y)~(\because~z=-x-y)$$ $$=\frac{(-x\cdot y)[(|y|^2+x\cdot y)(|x|^2+x\cdot y)-(|x|^2|y|^2-(x\cdot y)^2)]-(|x|^2|y|^2-(x\cdot y)^2)|z|^2}{|x|^2|y|^2|z|^2}$$ $$=\frac{-(x\cdot y)^2(| x|^2+|y|^2+2x\cdot y)-|x|^2|y|^2|z|^2+(x\cdot y)^2|z|^2}{| x|^2|y|^2|z|^2}=-1,$$ since by (1) the first and the third terms in the numerator cancel. QED