In $\Delta ABC$ if $a^2+b^2+c^2-ac-\sqrt{3}ab=0$ Then Prove that Triangle is Right angled

Question

In $\Delta ABC$ if $$a^2+b^2+c^2-ac-\sqrt{3}ab=0$$ Then Prove that Triangle is Right angled

I tried in this way:

Multiplying with $2$ both sides we get

$$2a^2+2b^2+2c^2-2ac-2\sqrt{3}ab=0$$

$$(a-c)^2+(a-\sqrt{3}b)^2+c^2=b^2$$

Also

$$(a-c)^2+(b-\sqrt{3}a)^2+c^2=2a^2-b^2$$

How to proceed?