In  $\Delta ABC$ if $a^2+b^2+c^2=ca+ab\sqrt{3}$  then the triangle is

$\begin{array}{l}{a}^2+b^2+c^2=ca+ab\sqrt{3}\\ \Rightarrow a^2+b^2+c^2-ab\sqrt{3}-ca=0\end{array}$

$\Rightarrow \frac{a^2}{4}+\frac{3a^2}{4}+b^2+c^2-ab\sqrt{3}-ca=0$

$\Rightarrow \frac{a^2}{4}+c^2-ac+\frac{3a^2}{4}+b^2-\frac{\sqrt{3}}{2}.a2b=0$

$\Rightarrow {\left(\frac{a}{2}-c\right)}^2+{\left(\frac{\sqrt{3}a}{2}-b\right)}^2=0$

This equation is satisfied when $C=\frac{a}{2};b=\frac{\sqrt{3}a}{2}$

$\therefore b^2+c^2=\frac{3a^2}{4}+\frac{a^2}{4}$

$=a^2$

$\begin{array}{l}A=90\\ \therefore C=\frac{a}{2}\Rightarrow sinC=\frac{sin90}{2}\\ \Rightarrow sinC=\frac{1}{2}\\ \Rightarrow C=30\end{array}$