What is the least number of (fixed) parameters I can ask for, when calculating area of a triangle of unknown type? I need to calculate the area of a triangle, but I don't know, whether it is right angled, isoscele or equilateral.
What parameters do I need to calculate the area of a triangle of unknown type?

All you need is the lengths of each side of the triangle.
By Heron's Formula, we know for a triangle with sides $a,b,c$, we have
$$A=\sqrt{s(s-a)(s-b)(s-c)}\text{ ,where }s=\frac{a+b+c}2$$
Reference:
https://en.wikipedia.org/wiki/Heron%27s_formula

EDIT:
In response to suggestion by @Hurkyl , I now add the case of ASA and SsA.

(ASA)
With a known side $c$ and its neighbouring angles $\alpha,\beta$, we have:
$$A=\frac{c^2\sin\alpha\sin\beta}{2\sin{(\alpha+\beta)}}$$
Proof:
Let $\alpha,\beta,\theta$ be the corresponding angles of sides $a,b,c$ respectively, we have
$$\theta=\pi-\alpha-\beta\implies\sin\theta=\sin{(\pi-\alpha-\beta)}=\sin{(\alpha+\beta)}$$
By Sine Law,
$$\frac c{{\sin{(\pi-\alpha-\beta)}}}=\frac a{\sin\alpha}=\frac b{\sin\beta}$$
So $a=\frac{c\sin\alpha}{\sin{(\alpha+\beta)}}$, $b=\frac{c\sin\beta}{\sin{(\alpha+\beta)}}$
Therefore,
\begin{align}
A&=\frac12ab\sin\theta\\&=\frac12\cdot{\frac{c\sin\alpha}{\sin{(\alpha+\beta)}}}\cdot\frac{c\sin\beta}{\sin{(\alpha+\beta)}}\cdot\sin{(\alpha+\beta)}\\&=\frac{c^2\sin\alpha\sin\beta}{2\sin{(\alpha+\beta)}}
\end{align}

(SsA)
With 2 known sides $b,c$ and $\beta$, the corresponding angle of $b$, we have:
$$A=\frac12 c\sin\beta[\sqrt{1-(\frac{c\sin\beta}b)^2}+c\cos\beta]$$
Proof:
By Sine Law:
$$\frac b{\sin\beta}=\frac c{\sin\theta}$$
$$\sin\theta=\frac{c\sin\beta}b, \cos\theta=\sqrt{1-(\frac{c\sin\beta}b)^2}$$
By Cosine Law,
$$c^2=a^2+b^2-2ab\cos\theta$$
$$c^2-b^2+b^2\cos^2\theta=a^2-2ab\cos\theta+b^2\cos^2\theta$$
$$c^2-b^2\sin^2\theta=(a-b\cos\theta)^2$$
$$a=b\cos\theta+\sqrt{c^2-b^2\sin^2\theta}=b\cos\theta+c\cos\beta$$
So
\begin{align}
A&=\frac12ab\sin\theta\\&=\frac12(b\cos\theta+c\cos\beta)(b)\cdot\frac{c\sin\beta}b\\&=\frac12 c\sin\beta[\sqrt{1-(\frac{c\sin\beta}b)^2}+c\cos\beta]
\end{align}
