You just need to apply the Angle Bisector Theorem again. For completeness, I'll go through the whole argument.
Given $\triangle ABC$ with side-lengths $a = |\overrightarrow{BC}|$, $b = |\overrightarrow{CA}|$, $c := |\overrightarrow{AB}|$, suppose the angle bisector from $A$ meets the opposite side at $D$. You're correct that, by the Angle Bisector Theorem
$$\frac{|\overrightarrow{BD}|}{|\overrightarrow{DC}|} = \frac{|\overrightarrow{AB}|}{|\overrightarrow{AC}|} = \frac{c}{b} \tag{$\star$}$$
so that we can write:
$$D = \frac{B b + C c}{b+c}$$
Note also that $(\star)$ implies
$$\frac{|\overrightarrow{BD}|}{|\overrightarrow{BC}|} = \frac{c}{b+c} \qquad\text{so that}\qquad |\overrightarrow{BD}| = |\overrightarrow{BC}|\frac{c}{b+c} = \frac{ac}{b+c}$$
Now, here's your next step: Suppose the angle bisector from $B$ meets $\overline{AD}$ at $X$ (which we "know" is the incenter). Again by the Angle Bisector Theorem applied to $\triangle ABD$,
$$\frac{|\overrightarrow{AX}|}{|\overrightarrow{XD}|} = \frac{|\overrightarrow{BA}|}{|\overrightarrow{BD}|} = \frac{c}{ac/(b+c)} = \frac{b+c}{a}$$
and we can write
$$X = \frac{A a + D (b+c)}{a+b+c} = \frac{A a + B b + C c}{a+b+c}$$
Because this formula is symmetric in the elements of $\triangle ABC$, we conclude that $X$ is not merely where the angle bisector from $A$ meets the angle bisector from $B$, but where any two angle bisectors meet ... and, therefore, where all three bisectors meet (aka, the incenter). $\square$
- Find co-ordinates/position vectors for the vertices of the triangle using:
$$r_1 = r_2$$
$$r_1 = r_3$$
$$r_2 = r_3$$
If you have found the three vertices, I'll name them $A$, $B$ and $C$, then form the following vectors: $\overrightarrow{AB}=B-A$ and $\overrightarrow{AC}=C-A$. By doing this, you actually 'move' the triangle $\Delta ABC$ to one with a vertex in the origin $O$, namely $\Delta O(B-A)(C-A)$ - with identical area of course.
The area of the parallellogram spanned by these two vectors is given by the magnitude of their cross product. The area of the triangle is exactly half of the area of that parallellogram, so:
$$\mbox{Area(triangle)} = \frac{\left| \overrightarrow{AB}\times\overrightarrow{AC}\right|}{2}
= \frac{\left| \left(B-A\right)\times\left(C-A\right)\right|}{2}$$
You should find $\sqrt{19} \approx 4.36$.
Best Answer
Example of a non-right-angled triangle where all dot products of position vectors vanish: \begin{align} \vec a &= \hat i\\ \vec b &= \hat j\\ \vec c &= \hat k \end{align}
Example of a right-angled triangle where no products of position vectors are zero: \begin{align} \vec a &= \hat i\\ \vec b &= \hat i + \hat j\\ \vec c &= \hat i + \hat k \end{align}
Thus orthogonality of position vectors is neither necessary nor sufficient for a triangle to be right-angled.