Probability that 3 points in a plane form a triangle

There is no such thing as a uniform distribution on the plane. Without specifying how the points are chosen, the question is not properly stated. However, if the points are chosen independently from some continuous distribution (absolutely continuous with respect to Lebesgue measure), the probability of the third point lying exactly on the line through the first two is $0$.

Nothing can be said about this as long as nothing has been said about the distribution (justifying the comment of angryavian).

Expressions "at random" or "are chosen" do not speak for themselves because there is no natural uniform distribution on $\mathbb R^2$.

If the distribution is absolutely continuous wrt the Lebesgue measure (i.e. if the distribution has a PDF) then automatically the answer is $1$ because every line in the plane $\mathbb R^2$ has Lebesgue measure $0$ (which is probably what Kaj means to say).

So in that case for any fixed line the probability that the third point is chosen on it equals $0$.

This is similar to this probability "joke":

Given a bowl with 9 black balls and 1 white ball, what's the chance that you pick a white ball? $\frac{1}{2}$, either you pick it or you don't.

While there are indeed both $\infty$ points which are collinear and $\infty$ points which are non-collinear, they're not quite the same $\infty$, so $\frac{\infty}{\infty+\infty}\neq \frac{1}{2}$. (See also Hilbert's Hotel on different levels of infinity)

As a matter of fact, since for collinear points, the choice of $x$ fixes the choice of $y$, there is only one level of infinity. For the non-collinear points, however, there are two levels of infinity: Both $x$ and $y$ can take infinite values. Thus, $\frac{P(\textrm{collinear})}{P(\textrm{non-collinear})}=\frac{1}{\infty}$, which tends to zero. In other words, $P(\textrm{non-collinear})\approx 1$.