On the parabola $y = x^2$ lie three different points $P, Q$ and $R$. Their projections $P', Q'$ and $R'$ on the $x$-axis are equidistant and equal to $s$ , i.e. $| P'Q'| = | Q'R'| = s$. Determine the area of $\vartriangle PQR$ in terms of $s$
1. **Define the coordinates of the points \( P, Q, \) and \( R \):**
   Given that the projections \( P', Q', \) and \( R' \) on the \( x \)-axis are equidistant and equal to \( s \), we can set the coordinates of the points as follows:
   \[
   P = (a-s, (a-s)^2), \quad Q = (a, a^2), \quad R = (a+s, (a+s)^2)
   \]
   Here, \( a \) is the \( x \)-coordinate of \( Q \).

2. **Apply the Shoelace Theorem:**
   The Shoelace Theorem (or Gauss's area formula) for the area of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) is given by:
   \[
   \text{Area} = \frac{1}{2} \left| x_1y_2 + x_2y_3 + x_3y_1 - y_1x_2 - y_2x_3 - y_3x_1 \right|
   \]
   Substituting the coordinates of \( P, Q, \) and \( R \):
   \[
   \text{Area} = \frac{1}{2} \left| (a-s)a^2 + a(a+s)^2 + (a+s)(a-s)^2 - (a-s)^2a - a^2(a+s) - (a+s)^2(a-s) \right|
   \]

3. **Simplify the expression:**
   Let's expand and simplify each term:
   \[
   (a-s)a^2 = a^3 - sa^2
   \]
   \[
   a(a+s)^2 = a(a^2 + 2as + s^2) = a^3 + 2a^2s + as^2
   \]
   \[
   (a+s)(a-s)^2 = (a+s)(a^2 - 2as + s^2) = a^3 - 2a^2s + as^2 + a^2s - 2as^2 + s^3 = a^3 - a^2s - as^2 + s^3
   \]
   \[
   (a-s)^2a = (a^2 - 2as + s^2)a = a^3 - 2a^2s + as^2
   \]
   \[
   a^2(a+s) = a^3 + a^2s
   \]
   \[
   (a+s)^2(a-s) = (a^2 + 2as + s^2)(a-s) = a^3 - a^2s + 2a^2s - 2as^2 + as^2 - s^3 = a^3 + a^2s - as^2 - s^3
   \]

4. **Combine and simplify the terms:**
   \[
   \text{Area} = \frac{1}{2} \left| (a^3 - sa^2) + (a^3 + 2a^2s + as^2) + (a^3 - a^2s - as^2 + s^3) - (a^3 - 2a^2s + as^2) - (a^3 + a^2s) - (a^3 + a^2s - as^2 - s^3) \right|
   \]
   \[
   = \frac{1}{2} \left| a^3 - sa^2 + a^3 + 2a^2s + as^2 + a^3 - a^2s - as^2 + s^3 - a^3 + 2a^2s - as^2 - a^3 - a^2s - a^3 - a^2s + as^2 + s^3 \right|
   \]
   \[
   = \frac{1}{2} \left| 3a^3 + 2a^2s + s^3 - 2a^3 - 2a^2s - s^3 \right|
   \]
   \[
   = \frac{1}{2} \left| a^3 + s^3 \right|
   \]
   \[
   = \frac{1}{2} \left| s^3 \right|
   \]
   Since \( s \) is a positive distance, \( s^3 \) is positive:
   \[
   \text{Area} = \frac{1}{2} s^3
   \]

The final answer is \(\boxed{\frac{1}{2} s^3}\)