1. Since $\overrightarrow{AB} \cdot \overrightarrow{AC} = \overrightarrow{BA} \cdot \overrightarrow{BC}$, we have $bc \cos A = ac \cos B$, which implies $b \cos A = a \cos B$.

By the sine rule, we have $\sin B \cos A = \sin A \cos B$.

Therefore, $\sin (A - B) = 0$.

Since $-π < A - B < π$, we have $A - B = 0$, hence $A = B$.

2. Since $\overrightarrow{AB} \cdot \overrightarrow{AC} = 1$, we have $bc \cos A = 1$.

By the cosine rule, we have $bc \cdot \frac{b^2 + c^2 - a^2}{2bc} = 1$, which implies $b^2 + c^2 - a^2 = 2$.

From part 1, we know that $a = b$, hence $c^2 = 2$, thus $c = \sqrt{2}$.

3. Since $|\overrightarrow{AB} + \overrightarrow{AC}| = \sqrt{6}$, we have $|\overrightarrow{AB}|^2 + |\overrightarrow{AC}|^2 + 2|\overrightarrow{AB} \cdot \overrightarrow{AC}| = 6$, which implies $c^2 + b^2 + 2 = 6$.

Hence, $c^2 + b^2 = 4$.

Since $c^2 = 2$, we have $b^2 = 2$, thus $b = \sqrt{2}$.

Therefore, $\triangle ABC$ is an equilateral triangle.

Hence, the area of $\triangle ABC$ is $S_{\triangle ABC} = \frac{\sqrt{3}}{4} \cdot (\sqrt{2})^2 = \boxed{\frac{\sqrt{3}}{2}}$.