Solution:
Answer: the triangle $DEF$ with $FAE$ parallel to $BC$, $DBF$ parallel to $CA$ and $DCE$ parallel to $AB$.

Let $\alpha$ be the angle between planes $ABC$ and $PBC$. Let $h$ be the perpendicular distance from $H$ to the line $BC$, and let $h_A$ be the perpendicular distance from $A$ to the line $BC$. Then $PH = h \tan \alpha$, and the altitude from $A$ to $PBC$ is $h_A \sin \alpha$. Hence if $PH$ is shorter than the altitude from $A$ we require that $h < h_A \cos \alpha < h_A$. Similar arguments apply for $B$ and $C$. So if $PH$ is the shortest then $H$ lies within triangle $DEF$.

If $H$ does lie within $DEF$, then if we make $\alpha$ sufficiently small we will have $h < h_A \cos \alpha$ and hence $PH$ will be shorter than the altitude from $A$. Similarly we can make $PH$ sufficiently short that $PH$ is less than the altitudes from $B$ and $C$. Hence the inside of $DEF$ is the required locus.