$\begin{aligned} & \text {} \because A, B, C \text { are in A.P } \\ & \therefore 2 B=A+C \\ & \text { Now, } A+B+C=180 \Rightarrow 3 B=180 \Rightarrow B=60^{\circ} \\ & \frac{c}{a} \sin 2 A+\frac{a}{c} \sin 2 C=\frac{c}{a} \times 2 \sin A \cos A+\frac{a}{c} \times 2 \sin C \cos C \\ & =2\left[\frac{c}{a} \times \frac{2 \Delta}{b c} \times \frac{\left(b^2+c^2-a^2\right)}{2 b c}+\frac{a}{c} \times \frac{2 \Delta}{a b} \times \frac{a^2+b^2-c^2}{2 a b}\right]\end{aligned}$
$\begin{aligned} & =\frac{2 \Delta}{a b^2 c}\left[b^2+c^2-a^2+a^2+b^2-c^2\right] \\ & =\frac{4 \Delta}{a c}=2 \times \sin B \\ & =2 \times \sin \left(60^{\circ}\right)=2 \times \frac{\sqrt{3}}{2}=\sqrt{3} \\ & \therefore\left(\frac{c}{a} \sin 2 A+\frac{a}{c} \sin 2 C\right)=\sqrt{3}\end{aligned}$