Area of a hexagon from the distances of the opposite sides
The general problem is not easy at all: in some cases, it is best to take the problem poser point of view. How would you construct a simple equilateral hexagon with parallel opposite sides? My inspired guess was to exploit the $(3,4,5)$ Pythagorean triple:
and I was so lucky that I met the exact constraints on the distances of opposite sides.
From the above diagram it is trivial to get that the side length is $5$ and the wanted area is
$$ 25+25-2+6+6 = \frac{5}{2}(7+8+9) = \color{red}{60}.$$
Similar to Jack D'Aurizo's suggestion, consider embedding our hexagon into $\mathbb{C}$ by placing one vertex at $z=0$, another at $z=r$, and a third at $z=re^{i\alpha}$, where $r$ is the common length of the six sides.
The leftmost vertex in the above figure is at $z=re^{i\alpha}$, so the vertex that forms the left endpoint of the top horizontal side is at $z=re^{i\alpha}+re^{i\pi-\beta}$, since the top-left side must be parallel to the bottom-right side. The imaginary part of this vertex will give us the distance from the bottom horizontal side to the top horizontal side, which we call $d_1$. So we have \begin{equation} d_1 = \Im(r(e^{i\alpha}+e^{i(\pi-\beta)})) = r(\sin\alpha + \sin(\pi-\beta)) = r(\sin\alpha+\sin\beta). \end{equation} Denoting the distance from the bottom-left side to the top-right side by $d_2$ and the distance from the top-left side to the bottom-right side by $d_3$, we similarly see that \begin{equation} d_2 = r(\sin\alpha+\sin\gamma) \quad\text{and}\quad d_3 = r(\sin\beta+\sin\gamma). \end{equation} (Edit: We didn't have to embed into $\mathbb{C}$ to get these equations. We could also have drawn a horizontal line through the leftmost vertex and vertical lines through the other two left-hand vertices, then computed the heights of the two resulting triangles.) Now we can start getting tricky. Since we expect $d_1,d_2$, and $d_3$ to determine many (perhaps all) of the properties of our hexagon, let's compute the angles from these distances. We have \begin{align*} d_1+d_2-d_3 &= 2r\sin\alpha\\ d_1-d_2+d_3 &= 2r\sin\beta\\ -d_1+d_2+d_3 &= 2r\sin\gamma, \end{align*} so \begin{align*} \alpha &= \arcsin\left(\frac{d_1+d_2-d_3}{2r}\right)\\ \beta &= \arcsin\left(\frac{d_1-d_2+d_3}{2r}\right)\\ \gamma &= \arcsin\left(\frac{-d_1+d_2+d_3}{2r}\right). \end{align*} Because the opposite sides of our hexagon are parallel, opposite angles must agree. This means that $4\pi=2\alpha+2\beta+2\gamma$, so \begin{equation} 2\pi = \arcsin\left(\frac{d_1+d_2-d_3}{2r}\right) + \arcsin\left(\frac{d_1-d_2+d_3}{2r}\right) + \arcsin\left(\frac{-d_1+d_2+d_3}{2r}\right). \end{equation} We can get some traction here by using the following addition formula: \begin{align*} \arcsin x &+ \arcsin y + \arcsin z\\ &= \arcsin\left(x\sqrt{1-y^2}\sqrt{1-z^2} + y\sqrt{1-x^2}\sqrt{1-z^2} + z\sqrt{1-x^2}\sqrt{1-y^2}-xyz\right). \end{align*} Applying this to our above equation and multiplying through by $8r^3$ yields \begin{align*} 0 = (d_1 &+d_2 - d_3)\sqrt{4r^2-(d_1-d_2+d_3)^2}\sqrt{4r^2-(-d_1+d_2+d_3)^2}\\ &+(d_1 - d_2 + d_3)\sqrt{4r^2-(d_1+d_2-d_3)^2}\sqrt{4r^2-(-d_1+d_2+d_3)^2}\\ &+(-d_1 + d_2 + d_3)\sqrt{4r^2-(d_1+d_2-d_3)^2}\sqrt{4r^2-(d_1d_2+d_3)^2}\\ &-(d_1+d_2-d_3)(d_1-d_2+d_3)(-d_1+d_2+d_3). \end{align*} This equation can be rearranged to produce \begin{align*} 1&= \frac{\sqrt{4r^2-(d_1-d_2+d_3)^2}}{d_1-d_2+d_3}\frac{\sqrt{4r^2-(-d_1+d_2+d_3)^2}}{-d_1+d_2+d_3} + \frac{\sqrt{4r^2-(d_1+d_2-d_3)^2}}{d_1+d_2-d_3}\frac{\sqrt{4r^2-(-d_1+d_2+d_3)^2}}{-d_1+d_2+d_3} + \frac{\sqrt{4r^2-(d_1+d_2-d_3)^2}}{d_1+d_2-d_3}\frac{\sqrt{4r^2-(d_1-d_2+d_3)^2}}{d_1-d_2+d_3}. \end{align*} This seems like a good place to hand things over to a computer algebra system. I asked Mathematica to solve \begin{equation} 1 = \frac{\sqrt{4r^2-x^2}}{x}\frac{\sqrt{4r^2-y^2}}{y}+\frac{\sqrt{4r^2-x^2}}{x}\frac{\sqrt{4r^2-z^2}}{z} + \frac{\sqrt{4r^2-y^2}}{y}\frac{\sqrt{4r^2-z^2}}{z} \end{equation} for $r$ in terms of $x$, $y$, and $z$. The result was \begin{equation} r = \frac{xyz}{\sqrt{-x^4+2x^2y^2-y^4+2x^2z^2+2y^2z^2-z^4}} = \frac{xyz}{\sqrt{(x+y+z)(-x+y+z)(x-y+z)(x+y-z)}}. \end{equation} We can then make the substitutions $x=d_1+d_2-d_3$, $y=d_1-d_2+d_3$, and $z=-d_1+d_2+d_3$ to obtain the following expression for $r$ in terms of the distances: \begin{equation} \boxed{r = \frac{(-d_1 + d_2 + d_3) (d_1 + d_2 - d_3) (d_1 - d_2 + d_3)}{\sqrt{(d_1+d_2+d_3)(3d_1-d_2-d_3)(3d_2-d_3-d_1)(3d_3-d_1-d_2)}}.} \end{equation} (Thanks to Blue for simplifying these expressions!) The area of the hexagon will thus be \begin{equation} \boxed{A = \frac{1}{2}r(d_1+d_2+d_3) = \frac{(-d_1 + d_2 + d_3) (d_1 + d_2 - d_3) (d_1 - d_2 + d_3)\sqrt{d_1+d_2+d_3}}{2\sqrt{(3d_1-d_2-d_3)(3d_2-d_3-d_1)(3d_3-d_1-d_2)}}.} \end{equation} It's worth pointing out that in our case we have $d_1=9$, $d_2=8$, and $d_3=7$ (up to some permutation of labels), so \begin{equation} r = \frac{(6)(8)(10)}{\sqrt{(24)(12)(8)(4)}} = \frac{480}{\sqrt{9216}} = \frac{480}{96} = 5, \end{equation} as concluded by Jack. This solution is not nearly as inspired or pleasant to look at as Jack's, but I think it provides the generality you hoped for.
Edit. Here's some of the work we asked Mathematica to do. Let's rewrite the equation as \begin{equation} 1 = \sqrt{\left[4\left(\frac{r}{x}\right)^2-1\right]\left[4\left(\frac{r}{y}\right)^2-1\right]} + \sqrt{\left[4\left(\frac{r}{x}\right)^2-1\right]\left[4\left(\frac{r}{z}\right)^2-1\right]} + \sqrt{\left[4\left(\frac{r}{y}\right)^2-1\right]\left[4\left(\frac{r}{z}\right)^2-1\right]}. \end{equation} We can move the rightmost root to the LHS and square to obtain \begin{equation} \left(1 - \sqrt{\left[4\left(\frac{r}{y}\right)^2-1\right]\left[4\left(\frac{r}{z}\right)^2-1\right]}\right)^2 = \left[4\left(\frac{r}{x}\right)^2-1\right]\left(\sqrt{\left[4\left(\frac{r}{x}\right)^2-1\right]\left[4\left(\frac{r}{y}\right)^2-1\right]} + \sqrt{\left[4\left(\frac{r}{x}\right)^2-1\right]\left[4\left(\frac{r}{z}\right)^2-1\right]}\right)^2. \end{equation} Solve this for $4(r/x)^2$ and you'll find that \begin{equation} \frac{4x^2r^2}{y^2z^2} = 4\left(\frac{r}{y}\right)^2 + 4\left(\frac{r}{z}\right)^2 + 2\sqrt{4\left(\frac{r}{y}\right)^2-1}\sqrt{4\left(\frac{r}{z}\right)^2-1} - 2. \end{equation} Now isolate the remaining radicals and square again. After factoring out $\frac{4r^2}{y^4z^4}$ you'll have an expression from which we can easily find $r^2$.
Extend every other edge of the hexagon (of side-length $s$) to determine three points $A$, $B$, $C$; define $a := |BC|$, $b := |CA|$, $c := |AB|$.
Note that the parallel edges of the hexagon give rise to similar triangles. The figure labels segments according to the appropriate proportions.
Now, consider the hexagon's opposite-edge distance $k$. By similarity, we can write (with $T := |\triangle ABC|$)
$$\frac{|BB_c|}{|BC|} = \frac{k}{\text{altitude from $C$}} \quad\to\quad \frac{s(a+b)}{ab} = \frac{k}{2T/c} \quad\to\quad a b ck=2Ts(a+b) \tag{1} $$ Likewise, $$a b c h = 2 T s (b+c) \qquad a b c j = 2 T s (c+a) \tag{2}$$
Thus,
$$h : j : k \;=\; b+c:c+a:a+b \tag{3}$$
and we deduce that, for some $\lambda$,
$$a = (-h + j + k)\lambda \qquad b = (h-j+k)\lambda \qquad c = (h+j-k)\lambda \tag{4}$$ so that, by Heron's Formula, $$\begin{align} T &=\frac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)} \\[4pt] &= \frac{\lambda^2}{4}\sqrt{(h+j+k)(3h-j-k)(3j-k-h)(3k-h-j)} \end{align} \tag{5}$$ Finally, substituting $(4)$ and $(5)$ back into $(1)$:
$$s =\frac{(-h+j+k)(h-j+k)(h+j-k)}{\sqrt{(h+j+k)(3h-j-k)(3j-k-h)(3k-h-j)}} \tag{6}$$
The hexagon's area follows immediately. $\square$