1. Given that $p$, $q$, and $r$ are real numbers, and $r+q=6-4p+$ $3p^{2}$, $r-q=5-4p+p^{2}$. Then the relationship in size between $p$, $q$, and $r$ is ( ).
(A) $r<p<q$
(B) $q<r<p$
(C) $q<p<r$
(D) $p<q<r$
$-1 .(\mathrm{D})$
Completing the square, we get $r-q=(p-2)^{2}+1 \geqslant 1>0$, which implies $q0$, we know $q>p$.