Let be the origin, the point A be $$z=\sqrt{3}+2\sqrt{2}i$$, the point $$B(z_{2})$$ be such that $$\sqrt{3}|z_{2}|=|z_{1}|$$ and $$arg(z_{2})=arg(z_{1})+\frac{\pi}{6}$$. Then

We denote O as the origin, A by $$z_1 = \sqrt{3} + 2\sqrt{2}\,i$$, and B by $$z_2$$, where $$\sqrt{3}\,|z_2| = |z_1|$$ and $$\arg(z_2) = \arg(z_1) + \frac{\pi}{6}$$, and we aim to identify the correct statement about triangle ABO.

Since $$|z_1| = \sqrt{(\sqrt{3})^2 + (2\sqrt{2})^2} = \sqrt{3 + 8} = \sqrt{11}$$, it follows from $$\sqrt{3}\,|z_2| = \sqrt{11}$$ that $$|z_2| = \frac{\sqrt{11}}{\sqrt{3}} = \sqrt{\frac{11}{3}}\,. $$

We write $$\arg(z_1) = \tan^{-1}\!\bigl(\tfrac{2\sqrt{2}}{\sqrt{3}}\bigr) = \theta$$ (say), so that $$z_2 = |z_2|e^{i(\theta + \pi/6)} = \sqrt{\frac{11}{3}}\;e^{i(\theta + \pi/6)}\,. $$

Noting that $$OA = |z_1| = \sqrt{11}$$ and $$OB = |z_2| = \sqrt{\frac{11}{3}}\,, $$ we observe immediately that $$OA \neq OB$$, so triangle ABO is not isosceles in the sides OA and OB.

In order to compare AB and OB, we compute

$$|AB|^2 = |z_1 - z_2|^2 = |z_1|^2 + |z_2|^2 - 2\,|z_1|\,|z_2|\cos\Bigl(\tfrac{\pi}{6}\Bigr) = 11 + \tfrac{11}{3} - 2\sqrt{11}\,\sqrt{\tfrac{11}{3}}\;\frac{\sqrt{3}}{2} = \frac{11}{3}\,, $$

which gives $$|AB| = \sqrt{\frac{11}{3}} = |z_2| = OB\,. $$ Hence AB = OB, and triangle ABO is isosceles with AB = OB.

To determine whether it is obtuse, we apply the cosine rule at B opposite the longest side OA. Since

$$OA^2 = OB^2 + AB^2 - 2\,OB\,AB\cos B$$

we have

$$11 = \tfrac{11}{3} + \tfrac{11}{3} - 2\cdot\tfrac{11}{3}\cos B \quad\Longrightarrow\quad \cos B = -\tfrac{1}{2}\,, $$

so that $$B = 120^\circ > 90^\circ$$ and triangle ABO is obtuse.

Triangle ABO is an obtuse angled isosceles triangle, which corresponds to Option 2.