The condition for equations $\overrightarrow{\mathrm{r}}\times \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{b}}\text{ and }\overrightarrow{\mathrm{r}}\times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{d}}$ to be consistent is

$\begin{array}{l}\overrightarrow{\mathrm{r}}\times \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{b}}\\ \text{ or }\overrightarrow{\mathrm{d}}\times (\overrightarrow{\mathrm{r}}\times \overrightarrow{\mathrm{a}})=\overrightarrow{\mathrm{d}}\times \overrightarrow{\mathrm{b}}\end{array}$

$\begin{array}{l}\mathrm{or} (\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{d}})\overrightarrow{\mathrm{r}}-(\overrightarrow{\mathrm{d}}\cdot \overrightarrow{\mathrm{r}})\overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{d}}\times \overrightarrow{\mathrm{b}}----\left(\mathrm{i}\right)\\ \overrightarrow{\mathrm{r}}\times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{d}}\\ \mathrm{or} \overrightarrow{\mathrm{b}}\times (\overrightarrow{\mathrm{r}}\times \overrightarrow{\mathrm{c}})=\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{d}}\\ \mathrm{or} (\overrightarrow{\mathrm{b}}\cdot \overrightarrow{\mathrm{c}})\overrightarrow{\mathrm{r}}-(\overrightarrow{\mathrm{b}}\cdot \overrightarrow{\mathrm{r}})\overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{d}---\left(\mathrm{ii}\right)}\end{array}$

Adding (i) and (ii), we get 

$(\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{d}}+\overrightarrow{\mathrm{b}}\cdot \overrightarrow{\mathrm{c}})\overrightarrow{\mathrm{r}}-(\overrightarrow{\mathrm{d}}\cdot \overrightarrow{\mathrm{r}})\overrightarrow{\mathrm{a}}-(\overrightarrow{\mathrm{b}}\cdot \overrightarrow{\mathrm{r}})\overrightarrow{\mathrm{c}}=\overrightarrow{0}$

$\text{ Now }\overrightarrow{r}\cdot \overrightarrow{d}=0\text{ and }\overrightarrow{b}\cdot \overrightarrow{r}=0\text{ as }\overrightarrow{d}\text{ and }\overrightarrow{r}\text{ as well }$

$\text{ as }\overrightarrow{\mathrm{b}}\text{ and }\overrightarrow{\mathrm{r}}\text{ are mutually perpendicular. }$

Hence $(\overrightarrow{\mathrm{b}}\cdot \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{d}})\overrightarrow{\mathrm{r}}=\overrightarrow{0}$