Without loss of generality assume that equation of the parabola is $y = x^2$ (Fig. 2). Let the abscissas of the points $A$, $B$, and $C$ be $a$, $b$, and $c$. Let $A'$, $B'$, and $C'$ be the projections of $A$, $B$ and $C$ onto the $x$-axis. Denoting the area of a region $K$ by $S_K$ we have
$$S_{ABC} = S_{ACC'A'} - S_{ABB'A'} - S_{BCC'B'}.$$ Since $S_{ACC'A'} = 2d \cdot \frac{a^2+c^2}{2}$, $S_{ABB'A'} = d \cdot \frac{a^2+b^2}{2}$, $S_{BCC'B'} = d \cdot \frac{b^2+c^2}{2}$, it follows that

$$\frac{1}{2}d(2a^2 + 2c^2 - a^2 - b^2 - b^2 - c^2) = \frac{1}{2}d(a^2 - b^2 - b^2 + c^2) = \frac{1}{2}d(d(a+b) - d(b+c)) = \frac{1}{2}d^2(a-c) = \frac{1}{2}d^2 \cdot 2d = d^3.$$

Without loss of generality we can assume that point $B$ lies at the origin. Then the equation of the parabola is $y = x^2 + px$ with some $p$. The coordinates of $A$ and $C$ are then $(d, d^2 + pd)$ and $(-d, d^2 - pd)$. Let $D$ be the midpoint of the segment $AC$. Its coordinates are $(0, d^2)$, so $BD$ is perpendicular to the $x$-axis, hence the lengths of the altitudes of both triangles $ABD$ and $BCD$ with the base $BD$ are $d$. So both the triangles have area $\frac{d^3}{2}$, hence the area of the triangle $ABC$ is $d^3$.