<s>[INST] In a $\triangle ABC,a^2+b^2+c^2=ac+ab\sqrt3$,then the triangle is In a $\triangle ABC,a^2+b^2+c^2=ac+ab\sqrt3$,then the triangle is
$(A)$equilateral
$(B)$isosceles
$(C)$right angled
$(D)$none of these

The given condition is $a^2+b^2+c^2=ac+ab\sqrt3$.
Using sine rule,
$a=2R\sin A,b=2R\sin B,c=2R\sin C$,we get
$\sin^2A+\sin^2B+\sin^2C=\sin A\sin C+\sin A\sin B\sqrt3$

I am stuck here.
 [/INST] Considering cosine law:
\begin{align*}
  \sum_{abc} (b^2+c^2-a^2) &= \sum_{abc} 2bc\cos A \\
  a^2+b^2+c^2&= 2bc\cos A+2ca\cos B+2ab\cos C
\end{align*}
One possibility is $\cos A=0$, $2\cos B=1$ and $2\cos C=\sqrt{3}$.
Hence $A=90^{\circ}$, $B=60^{\circ}$ and $C=30^{\circ}$.

Proof of Uniqueness of the above solution:
Rearrange the equality as a quadratic in $b$:
$$b^2-(a\sqrt{3})b+(a^2-ac+c^2) =0 $$
The discriminant:
  \begin{align*}
  \Delta &= (-a\sqrt{3})^2-4(a^2-ac+c^2) \\
  &=-a^2+4ac-4c^2 \\
  &=-(a-2c)^{2} \\
  &\leq 0
\end{align*}
To admit a real solution for $b$, $\Delta$ has to be zero.  That is
$$a=2c$$
Now, $$3c^2-2bc\sqrt{3}+b^2=0 \implies b=c\sqrt{3}$$
Therefore, $$\fbox{$a^2=b^2+c^2$}$$

</s>