What you are looking for is called the shoelace formula:
\begin{align*}
\text{Area}
&= \frac12 \big| (x_A - x_C) (y_B - y_A) - (x_A - x_B) (y_C - y_A) \big|\\
&= \frac12 \big| x_A y_B + x_B y_C + x_C y_A - x_A y_C - x_C y_B - x_B y_A \big|\\
&= \frac12 \Big|\det \begin{bmatrix}
x_A & x_B & x_C \\
y_A & y_B & y_C \\
1 & 1 & 1
\end{bmatrix}\Big|
\end{align*}
The last version tells you how to generalize the formula to higher dimensions.
PS. Another generalization of the formula is obtained by noting that it follows from a discrete version of the Green's theorem:
$$ \text{Area} = \iint_{\text{domain}}dx\,dy = \frac12\oint_{\text{boundary}}x\,dy - y\,dx $$
Thus the signed (oriented) area of a polygon with $n$ vertexes $(x_i,y_i)$ is
$$ \frac12\sum_{i=0}^{n-1} x_i y_{i+1} - x_{i+1} y_i $$
where indexes are added modulo $n$.
You were on the right track.
Using your notation, $A = \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{(x + 3)xyz}$, but also $A = x + 3$. So
$$\begin{align}
\sqrt{(x + 3)xyz} &= x + 3, \\
xyz &= x + 3, \\
x &= \frac{3}{yz - 1}.
\end{align}$$
Substituting that into $A$ we get
$$A = \sqrt{\left(\frac{3}{yz - 1} + 3\right)\frac{3}{yz - 1}yz} = \sqrt{\frac{3yz}{yz - 1}\cdot\frac{3}{yz - 1}yz} = \frac{3yz}{yz - 1} = \frac{3}{yz - 1} + 3.$$
Finally, by AM-GM we get lower bound:
$$A = \frac{3}{yz - 1} + 3 \geqslant \frac{3}{\frac{(y + z)^2}{4} - 1} + 3 = \frac{3}{\frac{9}{4} - 1} + 3 = \frac{12}{5} + 3 = \frac{27}{5} = 5.4$$
Because it is AM-GM, equality is reached when $y = z = 1.5$
P.S. You may wonder, why for some values of $y$ and $z$ expression $\frac{3}{yz - 1}$ may become infinite or negative. Isn't it strange? Besides, I myself silently assumed that it is positive. And there's a reason for that.
Positive values of $\frac{3}{yz - 1}$ correspond to the situation you describe: a circle inscribe in a triangle.
When it becomes infinite two sides $AB$ and $AC$ become parallel.
And finally, when it's negative, your circle is no longer an incircle, it becomes excircle.
Best Answer
Study the above diagram which describes the problem.
From the calculations of the arc lengths, we have:
$$3 = rA, \qquad4 = rB, \qquad 5 = rC$$ $$3 + 4 + 5 = rA + rB + rC = r(A + B + C)$$ $$r = \frac{12}{A + B + C} = \frac{6}{\pi}$$
We also know that the angles $A,B,C$ are in proportions corresponding to the respective arc length proportions. So, $$A = \frac{6\pi}{12}, \qquad B = \frac{8\pi}{12}, \qquad C = \frac{10\pi}{12}$$
I believe it is easy to compute the area of the inscribed triangle from here: It is given by
$$\frac{1}{2}r^2\sin A + \frac{1}{2}r^2\sin B + \frac{1}{2}r^2\sin C$$