Let $\overrightarrow{\mathrm{a}}=2\widehat{\mathrm{i}}-7\widehat{\mathrm{j}}+5\widehat{\mathrm{k}},\overrightarrow{\mathrm{b}}=\widehat{\mathrm{i}}+\widehat{\mathrm{k}}\text{ and }\overrightarrow{\mathrm{c}}=\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}-3\widehat{\mathrm{k}}$ be three given vectors. If $\overrightarrow{\mathrm{r}}$is a vector such that $\overrightarrow{\mathrm{r}}\times \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{c}}\times \overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{r}}\cdot \overrightarrow{\mathrm{b}}=0,\text{ then }\left|\overrightarrow{\mathrm{r}}\right|$ then is equal to :

Give $\overrightarrow{\mathrm{r}}\times \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{c}}\times \overrightarrow{\mathrm{a}}$
$\begin{array}{l}\Rightarrow (\overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{c}})\times \overrightarrow{\mathrm{a}}=\overrightarrow{0}\Rightarrow \overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{c}}\text{ is parallel to }\overrightarrow{\mathrm{a}}\\ \Rightarrow \overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{c}}=\mathrm{\lambda }\overrightarrow{\mathrm{a}}\Rightarrow \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{c}}+\mathrm{\lambda }\overrightarrow{\mathrm{a}}\\ \therefore \overrightarrow{\mathrm{r}}=(1+2\mathrm{\lambda })\widehat{\mathrm{i}}+(2-7\mathrm{\lambda })\widehat{\mathrm{j}}+(-3+5\mathrm{\lambda })\widehat{\mathrm{k}}\end{array}$
also $\overrightarrow{\mathrm{r}}\cdot \overrightarrow{\mathrm{b}}=0$
$\begin{array}{l}\Rightarrow (1+2\mathrm{\lambda })\left(1\right)+(-3+5\mathrm{\lambda })\left(1\right)=0\\ \Rightarrow \mathrm{\lambda }=\frac{2}{7}\\ \therefore \overrightarrow{\mathrm{r}}=\frac{11}{7}\widehat{\mathrm{i}}-\frac{11}{7}\widehat{\mathrm{k}}\Rightarrow \left|\overrightarrow{\mathrm{r}}\right|=\frac{11}{7}\sqrt{2}\end{array}$