I found the following question in a book:
There exists a regular triangle $OAB$ which has edge-length $2$. Let $H, I, J$ be a foot of the perpendicular line drawn from a point $P$ in $OAB$ to the edge $OA, AB, BO$ respectively. Also, let $C, D, E$ be a circle whose diameter is $PH, PI, PJ$ respectively. Then, find the area of the region where $P$ can move such that the interiors of each $C, D, E$ lie within the interior of $OAB$.
This question is easy to solve. Then, I'm interested in the third-version of this question. Here is my question.
Question: There exists a regular tetrahedron $ABCD$. Let $H_A, H_B, H_C, H_D$ be a foot of the perpendicular line drawn from a point $P$ in $ABCD$ to the face $BCD, ACD, ABD, ABC$ respectively. Then, let $V_{12}$ be the volume of the region where $P$ can move such that $P$ satisfies $(1), (2)$. Also, let $V_{2}$ be the volume of the region where $P$ can move such that $P$ satisfies $(2)$. Then, find ${V_{12}}/{V_2}$. Here,
(1) The interiors of each sphere whose diameter is $PH_A, PH_B, PH_C, PH_D$ lie within the interior of $ABCD$.
(2) $PH_A\le PH_B\le PH_C\le PH_D$.
I've got the following : $\frac{PH_D}{PH_A}$ can be any value in $1 \le \frac{PH_D}{PH_A} \le 3$.
However, I'm facing difficulty. Could you show me how to get the result?