Let $ABC$ be an equilateral triangle. Let $P$ and $S$ be points on $AB$ and $AC$, respectively, and let $Q$ and $R$ be points on $BC$ such that $PQRS$ is a rectangle. If $PQ = \sqrt3 PS$ and the area of $PQRS$ is $28\sqrt3$, what is the length of $PC$?
1. Given that $ABC$ is an equilateral triangle, we know that all sides are equal, i.e., $AB = BC = CA$.
2. Let $P$ and $S$ be points on $AB$ and $AC$ respectively, and $Q$ and $R$ be points on $BC$ such that $PQRS$ is a rectangle.
3. Given $PQ = \sqrt{3} PS$ and the area of $PQRS$ is $28\sqrt{3}$, we need to find the length of $PC$.

First, let's denote:
- $PS = x$
- $PQ = \sqrt{3} x$

Since $PQRS$ is a rectangle, the area of $PQRS$ can be expressed as:
\[ \text{Area} = PQ \times PS = (\sqrt{3} x) \times x = \sqrt{3} x^2 \]

Given that the area is $28\sqrt{3}$:
\[ \sqrt{3} x^2 = 28\sqrt{3} \]
\[ x^2 = 28 \]
\[ x = 2\sqrt{7} \]

Thus, $PS = 2\sqrt{7}$ and $PQ = \sqrt{3} \times 2\sqrt{7} = 2\sqrt{21}$.

4. Since $P$ and $S$ are on $AB$ and $AC$ respectively, and $PQRS$ is a rectangle, $PA = PS = 2\sqrt{7}$ and $PB = SC = 2x = 4\sqrt{7}$.

5. To find $PC$, we use Stewart's Theorem. Stewart's Theorem states that for a triangle $ABC$ with a cevian $AD$:
\[ AB^2 \cdot DC + AC^2 \cdot BD = AD^2 \cdot BC + BD \cdot DC \cdot BC \]

In our case, $A = A$, $B = B$, $C = C$, $D = P$, $BD = PB = 4\sqrt{7}$, $DC = PC$, and $BC = a$ (side of the equilateral triangle).

Since $AB = BC = CA = a$, we have:
\[ a^2 \cdot 4\sqrt{7} + a^2 \cdot 2\sqrt{7} = PC^2 \cdot a + 4\sqrt{7} \cdot 2\sqrt{7} \cdot a \]

Simplifying:
\[ 4a^2 \sqrt{7} + 2a^2 \sqrt{7} = PC^2 \cdot a + 8a \cdot 7 \]
\[ 6a^2 \sqrt{7} = PC^2 \cdot a + 56a \]

Dividing both sides by $a$:
\[ 6a \sqrt{7} = PC^2 + 56 \]

Since $a = 2\sqrt{7}$:
\[ 6 \cdot 2\sqrt{7} \cdot \sqrt{7} = PC^2 + 56 \]
\[ 6 \cdot 2 \cdot 7 = PC^2 + 56 \]
\[ 84 = PC^2 + 56 \]
\[ PC^2 = 28 \]
\[ PC = \sqrt{28} = 2\sqrt{7} \]

The final answer is $\boxed{2\sqrt{7}}$.