Given, $\overrightarrow{r}\times \overrightarrow{a}=\overrightarrow{b}\times \overrightarrow{a}$

$\Rightarrow \left(\overrightarrow{r}-\overrightarrow{b}\right)\times \overrightarrow{a}=0$
$$\begin{gathered} \Rightarrow \overrightarrow{r}-\overrightarrow{b}=\lambda \overrightarrow{a} \\ \Rightarrow \overrightarrow{r}=\overrightarrow{b}+\lambda \overrightarrow{a} \end{gathered}$$
$$\begin{gathered} \therefore \overrightarrow{r}\cdot \overrightarrow{c}=\overrightarrow{c}\cdot \overrightarrow{b}+\lambda \left(\overrightarrow{a}\cdot \overrightarrow{c}\right)=0 \\ \Rightarrow \lambda =-\frac{\overrightarrow{b}\cdot \overrightarrow{c}}{\overrightarrow{a}\cdot \overrightarrow{c}}\dots \dots ..\left(\mathrm{i}\right) \end{gathered}$$
$\overrightarrow{r}\times \overrightarrow{b}=\left(\overrightarrow{b}+\lambda \overrightarrow{a}\right)\times \overrightarrow{b}=\lambda \left(\overrightarrow{a}\times \overrightarrow{b}\right)\dots \dots ..\left(\mathrm{i}\mathrm{i}\right)$
$\overrightarrow{a}\times \overrightarrow{r}=\overrightarrow{a}\times \left(\overrightarrow{b}+\lambda \overrightarrow{a}\right)=\overrightarrow{a}\times \overrightarrow{b}\dots \dots ..\left(\mathrm{i}\mathrm{i}\mathrm{i}\right)$
Now from $\left(\mathrm{ii}\right) \& \left(\mathrm{iii}\right)$,

$\left(\overrightarrow{a}\cdot \overrightarrow{c}\right)\left(\overrightarrow{r}\times \overrightarrow{b}\right)+\left(\overrightarrow{b}\cdot \overrightarrow{c}\right)\left(\overrightarrow{a}\times \overrightarrow{r}\right)$
$=\left(\overrightarrow{a}\cdot \overrightarrow{c}\right)\lambda \left(\overrightarrow{a}\times \overrightarrow{b}\right)+\left(\overrightarrow{b}\cdot \overrightarrow{c}\right)\left(\overrightarrow{a}\times \overrightarrow{b}\right)$
$=\left(\overrightarrow{a}\times \overrightarrow{b}\right)\left[\left(\overrightarrow{a}\cdot \overrightarrow{c}\right)\lambda +\left(\overrightarrow{b}\cdot \overrightarrow{c}\right)\right]=0$ (from $\left(\mathrm{i}\right)$)