value of expression having variables $p,q,r,x,y,z$ 
If $p,q,r,x,y,z$ are non zero real number such that
$px+qy+rz+\sqrt{(p^2+q^2+r^2)(x^2+y^2+z^2)}=0$
Then $\displaystyle \frac{py}{qx}+\frac{qz}{ry}+\frac{rx}{pz}$ is

what  try
$(px+qy+rz)^2=(p^2+q^2+r^2)(x^2+y^2+z^2)$
$p^2x^2+q^2y^2+r^2z^2+2pqxy+2qryz+2prxz=p^2x^2+p^2y^2+p^2z^2+q^2x^2+q^2y^2+q^2z^2+r^2x^2+r^2y^2+r^2z^2$
$2pqxy+2qryz+2prxz=p^2y^2+p^2z^2+q^2x^2+q^2z^2+r^2x^2+r^2y^2$
How do i solve it Help me please

Write $v=(p,q,r)$ and $w=(x,y,z)$. Then the given relation states
$$v\cdot w+|v||w|=0$$
But $v\cdot w=|v||w|\cos\theta$ where $\theta$ is the angle between them, so
$$|v||w|(\cos\theta+1)=0$$
Since none of the scalars are zero, we get $\cos\theta=-1$, so $v=kw$ for some nonzero $k\in\mathbb R$ and
$$\frac{py}{qx}+\frac{qz}{ry}+\frac{rx}{pz}=\frac{kxy}{kyx}+\frac{kyz}{kzy}+\frac{kzx}{kxz}=3$$
