Let $X$ be the collection of random variables taking values from $[0,1]$.
Let $U$ be the sub-collection of uniform random variables.
For each $f \in X$, let $\rho_f \in [0,1] \to \mathbb{R}\cup \{\infty\}$ be its PDF whenever it is defined.
In particular, if $f \in U$, then $\rho_f(t) = 1$ for $t \in [0,1]$.
Let $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ be 3 random points from $U^2$.
Let $T$ be the triangle formed from them and $A$ be its area. Introduce a bunch of variables $x_\ell, x_m, x_u, y_\ell, y_m, y_u$ such that
$$\begin{cases}
x_\ell \le x_m \le x_u,& \{ x_\ell, x_m, x_u \} = \{ x_1, x_2, x_3 \}\\
y_\ell \le y_m \le y_u,& \{ y_\ell, y_m, y_u \} = \{ y_1, y_2, y_3 \}\\
\end{cases}$$
Let $w = x_u - x_\ell$ and $h = y_u - y_\ell$, they are the width and height of the bounding box of $T$.
It is easy to check $w, h \in X$ and
$$\rho_w(t) = \rho_h(t) = 6t(1-t)\quad\text{ for } t \in [0,1]$$
From this, we can deduce $wh \in X$ with
$$\begin{align}
\rho_{wh}(\ell) &= \int_0^1 \int_0^1 \rho_w(s)\rho_h(t) \delta(st - \ell) ds dt
= \int_\ell^1 \rho_w(s)\rho_h\left(\frac{\ell}{s}\right) \frac{ds}{s}\\
&= 36\ell \left[ -(\ell+1)\log \ell + 2(\ell-1) \right]
\end{align}
$$
Let $\displaystyle\;u = \frac1w ( x_m - x_\ell)$, $\displaystyle\;v = \frac1h ( y_m - y_\ell )$ and $\eta$ be the ratio $\displaystyle\;\frac{2A}{wh}$.
Aside from events of probability zero ($w = 0$ or $h = 0$), $u, v$ and $\eta$ are well defined. Furthermore, $u, v$ behave as if they belongs to $U$.
Given any realization of $( x_\ell, y_\ell)$, $( x_m, y_m )$, $( x_u, y_u )$, there are $(3!)^2 = 36$ ways to assign the coordinates to the 3 vertices $(x_1, y_1)$, $(x_2,y_2)$ and $(x_3,y_3)$.
Case I - In $12$ of the ways, two of the vertices occupy an pair of opposite corners of the bounding box. One example is
$$
\begin{cases}
(x_1,y_1) &= (x_\ell,y_\ell),\\
(x_2,y_2) &= (x_m,y_m)\\
(x_3,y_3) &= (x_u, y_u)
\end{cases}
\quad\implies\quad
\eta = \left|\begin{matrix}
0 & 0 & 1\\
u & v & 1\\
1 & 1 & 1
\end{matrix}\right| = |u-v|$$
In these cases, $\eta$ behaves like a random variable in $X$ with
$$\rho_{\eta,I}(t) = 2(1-t)$$
Case II - For the remaining $24$ ways, only one vertices is occupying an corner of the bounding box and the remaining two vertices lying on edges of the bounding box. One example is
$$
\begin{cases}
(x_1,y_1) &= (x_\ell,y_\ell),\\
(x_2,y_2) &= (x_u, y_m),\\
(x_3,y_3) &= (x_m, y_u)
\end{cases}
\quad\implies\quad
\eta = \left|\begin{matrix}
0 & 0 & 1\\
1 & v & 1\\
u & 1 & 1
\end{matrix}\right| = 1-uv$$
In these cases, $\eta$ behaves like a random variable $\in X$ with
$$\rho_{\eta,II}(t) = \rho_{uv}(1-t) = \int_{1-t}^1 \frac{ds}{s} = -\log(1-t)$$
Combine these two scenarios, we find aside from events of probability zero, we can treat $\eta$ as a random variable in $X$ with
$$\rho(t) = \frac13 \rho_{\eta,I}(t) + \frac23 \rho_{\eta,II}(t)
= \frac23\left[(1-t) - \log(1-t)\right]$$
This implies we can treat $2A = \eta wh$ as a random variable $\in X$ with PDF
$$\rho_{2A}(t)
=
\int_{t}^1 \rho_{wh}(s)\rho_\eta\left(\frac{t}{s}\right)\frac{ds}{s}$$
With help of a CAS, this evaluates to an ugly mess:
$$6(1-t)(1 - (5t+1)\log(1-t))
+ 6t^2\log(t)(\log(t) - 5)
+ 12t(t+2)\left(\mathrm{Li}_2(t) - \frac{\pi^2}{6}\right)
$$
where
$$\mathrm{Li}_2(z) =
- \int_0^z \frac{\log(1-t)}{t} dt = -\int_0^1 \frac{\log(1-zt)}{t} dt$$
is the Dilogarithm function.
As a double check, one can use above expression of $\rho_{2A}(t)$ to compute the expected area of $T$:
$$\mathbb{E}[A] = \frac12 \int_0^1 \rho_{2A}(t) t dt$$
This gives us an expected area $\frac{11}{144}$ matching what is in other answer.
