1. Identify the key points and their properties:
- $P$ and $R$ are the feet of the perpendiculars from $A$ onto the external and internal bisectors of $\angle ABC$, respectively.
- $Q$ and $S$ are the feet of the perpendiculars from $A$ onto the internal and external bisectors of $\angle ACB$, respectively.
- Given: $PQ = 7$, $QR = 6$, and $RS = 8$.

2. Establish collinearity and distances:
- Points $P$, $Q$, $R$, and $S$ lie on the $A$-midline of $\triangle ABC$, making them collinear.
- Since $APBR$ and $AQCS$ are rectangles, we have:
$$AB = PR = PQ + QR = 7 + 6 = 13$$
$$AC = QS = QR + RS = 6 + 8 = 14$$

3. **Reflect $A$ across $Q$ and $R$:**
- Let $Q'$ and $R'$ be the reflections of $A$ across $Q$ and $R$, respectively.
- The length $BC$ can be expressed as:
$$BC = BR' + CQ' - Q'R'$$
- Since $BR' = BA$ and $CQ' = CA$, we have:
$$BR' = AB = 13$$
$$CQ' = AC = 14$$
- The distance $Q'R'$ is twice the distance $QR$:
$$Q'R' = 2 \cdot QR = 2 \cdot 6 = 12$$

4. **Calculate $BC$:**
- Substitute the values into the equation for $BC$:
$$BC = BR' + CQ' - Q'R' = 13 + 14 - 12 = 15$$

5. **Use Heron's formula to find the area of $\triangle ABC$:**
- The semi-perimeter $s$ of $\triangle ABC$ is:
$$s = \frac{AB + AC + BC}{2} = \frac{13 + 14 + 15}{2} = 21$$
- Using Heron's formula:
$$\text{Area} = \sqrt{s(s - AB)(s - AC)(s - BC)} = \sqrt{21(21 - 13)(21 - 14)(21 - 15)}$$
$$= \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = \sqrt{21 \cdot 336} = \sqrt{7056} = 84$$

The final answer is $\boxed{84}$