Personally, I will think with vectors. It will end up at the same place nonetheless. Assuming you are finding coordinates of $D$.
We have $$\left|\overrightarrow{AD}\right|=\left|\overrightarrow{BD}\right|=\left|\overrightarrow{CD}\right|$$
with $\overrightarrow{OA}=\binom{100}{42}$, $\overrightarrow{OB}=\binom{33}{74}$, and $\overrightarrow{OC}=\binom{-26}{6}$.
Let $\overrightarrow{OC}=\binom{x}{y}$.
We have
$$(x-100)^2+(y-42)^2=(x-33)^2+(y-74)^2=(x+26)^2+(y-6)^2$$
This seems very complicated but it will be simpler if we do two at a time.
Simplifying pairwise, we have
$$y= \frac{67}{32}x-\frac{5199}{64}$$
$$y= -\frac{59}{68}x+\frac{5853}{136}$$
$$y= -\frac{7}{2}x-\frac{307}{2}$$
In fact, solving any two of the linear equations will give the result if the question is valid.
So we have
$$D\equiv\left(\frac{15023}{358},\frac{4745}{716}\right)$$
which, to be honest, is an unnecessarily ugly answer.
Diagram:
