To solve this problem, we need to consider the geometric properties of the triangle $\triangle ABC$ and the point $P$ inside it.

We are given that $D$, $E$, and $F$ are the feet of the perpendiculars from the point $P$ to the lines $BC$, $CA$, and $AB$, respectively. Our goal is to find the point $P$ such that the expression
$$S = \frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF}$$
is minimized.

### Analysis
For any point $P$ inside the triangle, the perpendicular distances from $P$ to the sides are $PD$, $PE$, and $PF$. The expression for minimizing involves the reciprocals of these distances, weighted by the side lengths opposite to each respective distance.

A notable point inside a triangle that often minimizes or optimizes such conditions is the Fermat Point (also known as the Torricelli point), which minimizes the total distance from the point to the vertices of the triangle. However, in this problem, the condition involves distances to sides, weighted by the lengths of those sides.

### Solution
Given the parallels with known geometric properties, it turns out that the incenter of the triangle $\triangle ABC$, denoted as $I$, can often split triangle-related expressions in a symmetric or optimizing way due to the nature of its equidistant properties to $AB$, $BC$, and $CA$.

To justify that $P = I$ minimizes $S$:
1. The incenter $I$ is equidistant to the sides due to being the intersection of angle bisectors.
2. By properties of reflections and symmetry in positive length weighting, dividing the sum by the respective perpendiculars counterbalances the weight on the side lengths, akin to finding a balance point or centroid-like behaviour (but with the unique symmetry that the incenter offers).

Hence, the sum $\frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF}$ is minimized when $P$ is the incenter $I$ of triangle $\triangle ABC$.

Thus, the point $P$ that minimizes the sum is:
$$\boxed{\text{The incenter of } \triangle ABC}.$$

Note: This results rely on the properties of geometric weights and known minimizing behaviours of the incenter. Further geometric proofs and investigations into this specific setup provide deeper validation.