1. Given point $A$

   $z_1=\sqrt{3}+2\sqrt{2}\,i$

   So its modulus is $|z_1|=\sqrt{(\sqrt{3})^2+(2\sqrt{2})^2}=\sqrt{3+8}=\sqrt{11}.$

   Also, $\tan\arg(z_1)=\frac{2\sqrt{2}}{\sqrt{3}}.$ Since both real and imaginary parts are positive, $A$ lies in the first quadrant.

2. Find modulus and argument of $z_2$

   We are given $\sqrt{3}|z_2|=|z_1|=\sqrt{11}$ so $|z_2|=\frac{\sqrt{11}}{\sqrt{3}}.$

   Also, $\arg(z_2)=\arg(z_1)+\frac{\pi}{6}.$

3. Lengths of sides of triangle $ABO$

   Since $O$ is origin:

   • $OA=|z_1|=\sqrt{11},$
   • $OB=|z_2|=\frac{\sqrt{11}}{\sqrt{3}}.$

   The angle between $OA$ and $OB$ is $\angle AOB=\arg(z_2)-\arg(z_1)=\frac{\pi}{6}.$

4. Find area of triangle $ABO$

   Using formula $\text{Area}=\frac12\, OA\cdot OB\cdot \sin\angle AOB,$ we get $\text{Area}=\frac12\cdot \sqrt{11}\cdot \frac{\sqrt{11}}{\sqrt{3}}\cdot \sin\frac{\pi}{6}.$

   Since $\sin \frac{\pi}{6}=\frac12$, $\text{Area}=\frac12\cdot \frac{11}{\sqrt{3}}\cdot \frac12=\frac{11}{4\sqrt{3}}.$

   So:

   • Option A: $\frac{11}{4}$ is false
   • Option B: $\frac{11}{\sqrt{3}}$ is false

5. Check whether triangle is scalene or isosceles

   We already have $OA=\sqrt{11},\qquad OB=\frac{\sqrt{11}}{\sqrt{3}}.$

   Now find $AB$ using cosine rule: $AB^2=OA^2+OB^2-2(OA)(OB)\cos\frac{\pi}{6}.$

   Substitute: $AB^2=11+\frac{11}{3}-2\cdot \sqrt{11}\cdot \frac{\sqrt{11}}{\sqrt{3}}\cdot \frac{\sqrt{3}}{2}.$

   Simplify the last term: $2\cdot \sqrt{11}\cdot \frac{\sqrt{11}}{\sqrt{3}}\cdot \frac{\sqrt{3}}{2}=11.$

   Hence $AB^2=11+\frac{11}{3}-11=\frac{11}{3},$ so $AB=\sqrt{\frac{11}{3}}=\frac{\sqrt{11}}{\sqrt{3}}=OB.$

   Therefore, $AB=OB,$ so triangle $ABO$ is isosceles, not scalene.

   Thus:

   • Option C is false.

6. Check whether the isosceles triangle is obtuse angled

   Since $AB=OB$, the equal sides meet at $B$, so the base is $OA$.

   The largest side is $OA=\sqrt{11}.$

   Check: $OA^2=11,$ and $OB^2+AB^2=\frac{11}{3}+\frac{11}{3}=\frac{22}{3}.$

   Since $11>\frac{22}{3},$ the angle opposite $OA$ is obtuse. That angle is $\angle ABO$.

   Hence triangle $ABO$ is an obtuse angled isosceles triangle.

   So option D is true.

7. Final evaluation of options

   • A: False
   • B: False
   • C: False
   • D: True

Therefore, the correct answer is D.