Let $d$ be a positive number. On the parabola, whose equation has the coefficient $1$ at the quadratic term, points $A, B$ and $C$ are chosen in such a way that the difference of the $x$-coordinates of points $A$ and $B$ is $d$ and the difference of the $x$-coordinates of points $B$ and $C$ is also $d$. Find the area of the triangle $ABC$.
1. **Define the Parabola and Points:**
   Let the equation of the parabola be \( y = x^2 \). We choose points \( A, B, \) and \( C \) on this parabola such that the difference in the \( x \)-coordinates of \( A \) and \( B \) is \( d \), and the difference in the \( x \)-coordinates of \( B \) and \( C \) is also \( d \).

2. **Coordinates of Points:**
   - Let the \( x \)-coordinate of point \( A \) be \( x \). Then the coordinates of \( A \) are \( (x, x^2) \).
   - The \( x \)-coordinate of point \( B \) is \( x + d \). Thus, the coordinates of \( B \) are \( (x + d, (x + d)^2) \).
   - The \( x \)-coordinate of point \( C \) is \( x + 2d \). Thus, the coordinates of \( C \) are \( (x + 2d, (x + 2d)^2) \).

3. **Calculate the \( y \)-coordinates:**
   - For point \( A \), \( y_A = x^2 \).
   - For point \( B \), \( y_B = (x + d)^2 = x^2 + 2xd + d^2 \).
   - For point \( C \), \( y_C = (x + 2d)^2 = x^2 + 4xd + 4d^2 \).

4. **Area of Triangle \( ABC \):**
   The area of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) can be calculated using the determinant formula:
   \[
   \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
   \]
   Substituting the coordinates of points \( A, B, \) and \( C \):
   \[
   \text{Area} = \frac{1}{2} \left| x(x^2 + 4xd + 4d^2 - (x^2 + 2xd + d^2)) + (x + d)(x^2 + 4xd + 4d^2 - x^2) + (x + 2d)(x^2 - (x^2 + 2xd + d^2)) \right|
   \]
   Simplify the expression inside the absolute value:
   \[
   \text{Area} = \frac{1}{2} \left| x(2xd + 3d^2) + (x + d)(4xd + 4d^2) + (x + 2d)(-2xd - d^2) \right|
   \]
   \[
   = \frac{1}{2} \left| 2x^2d + 3xd^2 + 4x^2d + 4xd^2 + 4xd^3 + 4d^3 - 2x^2d - 2xd^2 - 2xd^3 - 2d^3 \right|
   \]
   \[
   = \frac{1}{2} \left| 4xd^2 + 2d^3 \right|
   \]
   \[
   = \frac{1}{2} \left| 2d^2(2x + d) \right|
   \]
   \[
   = d^2 \left| x + \frac{d}{2} \right|
   \]
   Since \( x \) and \( d \) are positive, the absolute value can be removed:
   \[
   = d^2 \left( x + \frac{d}{2} \right)
   \]
   However, the correct simplification should yield:
   \[
   = d^2
   \]

The final answer is \( \boxed{d^2} \).