Let $\text{p},\text{ q},\text{ r}$  be positive real numbers, not all equal, such that some two of the equations  $p{x}^2+2qx+r=0,q{x}^2+2rx+p=0,r{x}^2+2px+q=0,$ have a common root, say $\alpha$,  then which of the following is/are the correct statement(s)  

Consider the discriminants of the three equations
$p{x}^2+qr+r=0$     (1)
$q{x}^2+rx+p=0$     (2)
  $r{x}^2+px+q=0$   (3)
 Let us denote them by  $D_1,D_2,D_3$ respectively. Then we have
  $D_1=4(q^2-rp),D_2=4(r^2-pq),D_3=4(p^2-qr).$
 We observe that
  $D_1+D_2+D_3=4(p^2+q^2+r^2-pq-qr-rp)$
 $=2\{{(p-q)}^2+{(q-r)}^2+{(r-p)}^2\}>0$
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-r eal roots and equation (1) has only real roots. Hence the common root  $\alpha$  must be between  (2) and (3). But then $\overline{\alpha }$  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 $D_2\ge 0$ . Thus, we have
    $q^2-rp>0,r^2-pq\ge 0$
These two gives
    $q^2{r}^2>p^{2}qr$
since p, q, r are all positive. Hence, we obtain $qr>p^2$  or  $D_3<0$. We conclude that the  common root must be between equations (1) and (2).
Thus 
  $p{\alpha }^2+q\alpha +r=0$
  $q{\alpha }^2+r\alpha +p=0$
Eliminating ${\alpha }^2$ , we obtain
 $2(q^2-pr)\alpha =p^2-qr$
Since  $q^2-pr>0$ and $p^2-qr<0$ , we conclude that $\alpha <0$ .
The condition   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  $\overline{\alpha }$ 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  $q^2\ge rp$ and $r^2\ge pq$ . Here at least one inequality is strict for $q^2=pr$   and $r^2=pq$  forces $p=q=r$ . Hence $q^2{r}^2>p^{2}qr$ . This gives  $p^2<qr$ and  hence (3) has nonreal roots.