Solution:
Consider the discriminants of the three equations
$$\begin{array}{r} p x^{2}+2 q x+r=0 \\ q x^{2}+2 r x+p=0 \\ r x^{2}+2 p x+q=0 \end{array}$$
Let us denote them by $D_{1}, D_{2}, D_{3}$ respectively. Then we have
$$D_{1}=4\left(q^{2}-r p\right), \quad D_{2}=4\left(r^{2}-p q\right), \quad D_{3}=4\left(p^{2}-q r\right)$$
We observe that
$$\begin{aligned} D_{1}+D_{2}+D_{3} & =4\left(p^{2}+q^{2}+r^{2}-p q-q r-r p\right) \\ & =2\left\{(p-q)^{2}+(q-r)^{2}+(r-p)^{2}\right\}>0 \end{aligned}$$
since $p, q, r$ are not all equal. Hence at least one of $D_{1}, D_{2}, D_{3}$ must be positive. We may assume $D_{1}>0$.
Suppose $D_{2}<0$ and $D_{3}<0$. In this case both the equations (2) and (3) have only non-real roots and equation (1) has only real roots. Hence the common root $\alpha$ must be between (2) and (3). But then $\bar{\alpha}$ is the other root of both (2) and (3). Hence it follows that (2) and (3) have same set of roots. This implies that
$$\frac{q}{r}=\frac{r}{p}=\frac{p}{q}$$
Thus $p=q=r$ contradicting the given condition. Hence both $D_{2}$ and $D_{3}$ cannot be negative. We may assume $D_{2} \geq 0$. Thus we have
$$q^{2}-r p>0, \quad r^{2}-p q \geq 0$$
These two give
$$q^{2} r^{2}>p^{2} q r$$
since $p, q, r$ are all positive. Hence we obtain $q r>p^{2}$ or $D_{3}<0$. We conclude that the common root must be between equations (1) and (2).
Thus
$$\begin{aligned} & p \alpha^{2}+2 q \alpha+r=0 \\ & q \alpha^{2}+2 r \alpha+p=0 \end{aligned}$$
Eliminating $\alpha^{2}$, we obtain
$$2\left(q^{2}-p r\right) \alpha=p^{2}-q r$$
Since $q^{2}-p r>0$ and $p^{2}-q r<0$, we conclude that $\alpha<0$.
The condition $p^{2}-q r<0$ implies that the equation (3) has only non-real roots.

Alternately one can argue as follows. Suppose $\alpha$ is a common root of two equations, say, (1) and (2). If $\alpha$ is non-real, then $\bar{\alpha}$ is also a root of both (1) and (2). Hence the coefficients of (1) and (2) are proportional. This forces $p=q=r$, a contradiction. Hence the common root between any two equations cannot be non-real. Looking at the coefficients, we conclude that the common root $\alpha$ must be negative. If (1) and (2) have common root $\alpha$, then $q^{2} \geq r p$ and $r^{2} \geq p q$. Here at least one inequality is strict for $q^{2}=p r$ and $r^{2}=p q$ forces $p=q=r$. Hence $q^{2} r^{2}>p^{2} q r$. This gives $p^{2}<q r$ and hence (3) has non-real roots.