Let $\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}=\hat{i}-\hat{j}+2 \hat{k}$ and $\vec{c}=5 \hat{i}-3 \hat{j}+3 \hat{k}$ be three vectors. If $\vec{r}$ is a vector such that, $\vec{r} \times \vec{b}=\vec{c} \times \vec{b}$ and $\vec{r} \cdot \vec{a}=0$, then $25|\vec{r}|^{2}$ is equal to :

$\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ <br/><br/>$\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$ <br/><br/>$\overrightarrow{\mathrm{c}}=\hat{5 \mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ <br/><br/>$(\overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{c}}) \times \vec{b}=0, \quad \vec{r} \cdot \vec{a}=0$ <br/><br/>$\Rightarrow \vec{r}-\vec{c}=\lambda \vec{b}$ <br/><br/>Also, $(\vec{c}+\lambda \vec{b}) \cdot \vec{a}=0$ <br/><br/>$\Rightarrow \vec{a} \cdot \vec{c}+\lambda(\vec{a} \cdot \vec{b})=0$ <br/><br/>$\therefore \lambda=\frac{\vec{a} \cdot \vec{c}}{\vec{a} \cdot \vec{b}}=\frac{-8}{5}$ <br/><br/>$\overrightarrow{\mathrm{r}}=\frac{5(5 \hat{\mathrm{i}}-3 \hat{\mathrm{i}}+3 \hat{\mathrm{k}})-8(\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})}{5}$ <br/><br/>$\Rightarrow$ $\overrightarrow{\mathrm{r}}=\frac{17 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}+\hat{\mathrm{k}}}{5}$ <br/><br/>$\Rightarrow$ $|\overrightarrow{\mathrm{r}}|^{2}=\frac{1}{25}(289+50)$ <br/><br/>$\Rightarrow$ $25|\overrightarrow{\mathrm{r}}|^{2}=339$